Extracting Elements From A Database Into a Select Form Field, & Printing Out Records

imbizsol

Hi all,

Not sure if this is the place to ask this, but I've been searching around for a few days here and not coming up with anything yet, so hopefully someone can point me in the right direction.

I have a database of approximately 300 companies. I need to create a drop down search menu so people can find records in a certain category. I want this menu to be able to pull the categories automatically from the database into the menu, then when they click on submit, the records/companies in that category are shown.

I found a couple of tutorials online on extracting database elements into the select form field, but the problem is when I click on submit, nothing happens.

Here are the two scripts/tutorials I tried, but they don't pull up any results for the categories that are pulled from the database, and I know there are records for those categories in the database. Am I missing the print results line somewhere?

#1: Here is the first tutorial I tried:
<HTML>
<HEAD>
<TITLE>Sample Form</TITLE>
</HEAD>
<BODY>

// create connection
$connection = mysql_connect("localhost","xxxxxx","xxxxxx")
or die("Couldn't make connection.");

// select database
$db = mysql_select_db("members", $connection)
or die("Couldn't select database.");

// create SQL statement
$sql = "SELECT category FROM members
ORDER BY category ASC";

// execute SQL query and get result
$sql_result = mysql_query($sql,$connection)
or die("Couldn't execute query.");

// put data into drop-down list box
while ($row = mysql_fetch_array($sql_result)) {

$category  = $row["category"];

$option_block .= "<OPTION value=\"$category\">$category</OPTION>";

}

<FORM method="post" action="test3.php">
<P>Category:<br>
<SELECT name="category">

<? echo "$option_block"; ?>

</SELECT>

</FORM>

</BODY>
</HTML>

#2: This is the second tutorial I tried:
<HTML>
<BODY>
<script language="php">
// create connection
$connection = mysql_connect("localhost","xxxx","xxxxx")
or die("Couldn't make connection.");

// select database
$db = mysql_select_db("members", $connection)
or die("Couldn't select database.");

if($react == "search_category") {
if($category) {
$query = "SELECT category FROM members ORDER BY category ASC";
$result = mysql_query($query, $connection);
if(mysql_num_rows($result)) {
print("<strong>$category</strong><p>");
}
} else {
print("<strong>no records are found, sorry.</strong><p>");
}
}
elseif ($react == "search_alpha") {
if(($company)) {
$query = "SELECT * FROM members ORDER BY company LIMIT " . $start . ", 5";
$result = mysql_query($query, $connection);
} else {
print("<strong>no records are found, sorry.</strong><p>");
}
}
else {
print("<center>Search the MHHP Members Database</center>");
}
</script>
<form method="POST" action="testnew.php"><div align="center"><center><p>Search Categories
<input type="hidden" name="react" value="search_record">
<select name="category" size="1">
<script language="php">
$query = "SELECT category FROM members ORDER BY category";
$result = mysql_query($query, $connection);
if(mysql_num_rows($result)) {
// we have at least one category, so show all categories as options in select form
while($row = mysql_fetch_row($result))
{
print("<option value=\"$row[0]\">$row[0]</option>");
}
} else {
print("<option value=\"\">No records were found, sorry</option>");
}
</script>
</select><input type="submit" value="Submit"></p></center></div>
</form>
<form method="POST" action="testnew.php"><div align="center"><center><p>Search Alphabetically
<input type="hidden" name="react" value="search_alpha">
<select name="category" size="1">
<script language="php">
$query = "SELECT company FROM members ORDER BY company";
$result = mysql_query($query, $connection);
if(mysql_num_rows($result)) {
// we have at least one category, so show all category as options in select form
while($row = mysql_fetch_row($result))
{
print("<option value=\"$row[0]\">$row[0]</option>");
}
} else {
print("<option value=\"\">No records were found, sorry</option>");
}
</script>
</select><input type="submit" value="Submit"></p></center></div>
</form>
</BODY>
</HTML>

Thanks in advance.

leatherback

Hi,

I looked at your code. It looks ok to me (Although here it is midnight, and I am a bit squared eyed at the moment ;-) )

The only thing I can see just now is:
You have <? echo "$option_block"; ?> in the first routine.

Have you tried:

<? echo $option_block ?>

Besides that: You might want to add the option distinct in your query, so you don't get any double options:
select distinct category FROM members ORDER BY category ASC

Good luck!

TodManPlaa

This may be of assistance - it is a function that I call from the form layout and it marks options as already selected if they are.

function CallExpenseDropdown($categoryAcc){
	/*make SQL statement*/
	$issuelist = "SELECT m.Category,m.mDescription FROM Members m ORDER BY m.Category ASC";
	$result = mysql_query($issuelist);
	/*Error check*/
	if(!$result){
		echo "There has been an error. \n";
		echo "Please copy the contents of this page and email it to Admin.<br>".$issuelist;
		print("errno: " . mysql_errno());
		print ("error: " . mysql_error());
	}
	/*start SELECT dropdown*/
	echo"<select name=\"category\">";
	/*Loop through values*/
	while ($myrow = mysql_fetch_array($result)) {
		$categoryAccount=$myrow["Category"];
		/*if the selected value is the same as the value passed to the function
		 then the value is marked as selected*/
		if ($categoryAccount==$categoryAcc) {
			printf("\t<option value=\"%s\" selected=\"selected\">%s - %s - %s</option>\n", 
			$myrow["Category"],$myrow["Description"],$myrow["Category"],$myrow["Description"]);
		}else{
			/*Else its a normal option*/
			printf("\t<option value=\"%s\">%s - %s - %s</option>\n", 
			$myrow["Category"],$myrow["Description"],$myrow["Category"],$myrow["Description"]);
		}
	}
	/*finish the slect dropdown*/
	echo "</select>";
}

I also agree with leatherback's point re you quoting the $option_block value - you don't need those quote marks.

Hope this helps

imbizsol

Thanks everyone for your responses to my question. I think I found the problem, but I haven't been able to make a solution work yet. My submit function isn't going anywhere.

the name of my file is called test.php, and the action of my form is going to test.php. Then my query and OPTION tags are as such:

$query = "SELECT category FROM members ORDER BY category";
$result = mysql_query($query, $connection);
if(mysql_num_rows($result)) {
// we have at least one category, so show all categories as options in select form
while($row = mysql_fetch_row($result))
{
print("<option value=\"$row[0]\">$row[0]</option>");
}
} else {
print("<option value=\"\">No records were found, sorry.</option>");
}

So when I view the source, I'm getting this as my option:

<option value="Some Company Inc.">Some Company Inc.</option>

Then when I click submit, it's just refreshing the page.

What I want to happen is that when a person clicks submit, it pulls up a listing of records for that particular category, and that's not happening yet. Do I need to change the option value to something else in order to get the records to pull up, or does the action command in the form need to go to another page?

TodManPlaa

So, is what you are saying that when the screen refreshes you want a different select list showing?

If that is the case, then you need something like the following (using snippets of your code):

if ($category){/*ie if the variable $category has arrived at the refreshed page*/
	$query = "SELECT category_key,record_key,whatever_records_you_want_displayed FROM member_records WHERE category_key=$category ORDER BY whatever_you_want"; 
	$result = mysql_query($query, $connection); 
	if(mysql_num_rows($result)) { 
		// we have at least one category, so show all categories as options in select form 
		echo"<select name=\"record\">";
		while($row = mysql_fetch_row($result)) { 
			print("<option value=\"$row[0]\">$row[0]</option>"); 
		} 
	} else { 
		print("<option value=\"\">No records were found, sorry.</option>"); 
	} 
}else{
	$query = "SELECT category FROM members ORDER BY category"; 
	$result = mysql_query($query, $connection); 
	if(mysql_num_rows($result)) { 
		// we have at least one category, so show all categories as options in select form 
		echo"<select name=\"category\">";
		while($row = mysql_fetch_row($result)) { 
			print("<option value=\"$row[0]\">$row[0]</option>"); 
		} 
	} else { 
		print("<option value=\"\">No records were found, sorry.</option>"); 
	} 
}
echo "</select>\n";

So the above will generate two versions of a select dropdown - if the user has not already selected a categoty, then they'll be shown the category list, else if they have already chosen a category they'll be shown the recors list.

I'm still not sure if this is what you're trying to do - if it isn't, post a few rows of sample information from your tables so I can try to get a handle on what you're trying to do.

Cheers

imbizsol

okay, I've been trying to get this search script working all week and the more I try, the more frustrated I get :-)

Here's the URL where I'm testing this script:

http://www.madisonheightschamber.org/testnew.php

There are two search boxes here. One is for searching the database by category and one is for alphabetical.

My database is a members database of almost 400 records. It consists of the following fields:

Category
Company
Contact
Address
City
State
Zip
Phone
Fax
Email
URL

When I finish importing the data into mySQL, I should have almost 400 companies listed in about 50 different categories. Right now, I only have three records in different categories in the database just to test this script.

What I'm trying to do for the first search (the category search) is have the script pull the categories from the database automatically and list them in the drop down box. Then when a person clicks on one of the categories (for instance, Automotive Services), they can hit the submit button and get a page with all the companies listed in the database that have Automotive Services listed as their category.

The resulting page after clicking submit should have each company record (company, contact, address, city, state, zip, phone, fax, email, and URL) listed in that specific category in alphabetical order.

Right now, the script only pulls the categories into the drop down form, but I'm not sure how to get a results page listing the company records for the selected category.

I hope that makes sense. You might understand better if you see the test script I have up right now. You can select a category, but when you click submit, nothing happens, no company records are shown.

TodManPlaa

okay, I've been trying to get this search script working all week and the more I try, the more frustrated I get :-)

Believe me I know how you feel!

What I'm trying to do for the first search (the category search) is have the script pull the categories from the database automatically and list them in the drop down box.

so that appears to be working already...

they can hit the submit button and get a page with all the companies listed in the database that have Automotive Services listed as their category.

and this is the problem bit right?

OK, so from what I understand, if I choose locksmiths for example, you don't want the resulting page to show a dropdown right - you just want it to show a table holding all the locksmiths?

I'll assume that's want you want to do - if not, let me know and I'll throw myself off the balcony.

That being the case - you need an if else structure on your page - something like this:

//top of page HTML here
if ($category){
	/*if this variable is present on the page, 
it means that the user has selected a particular category
from the form and hit submit
	so run a query to get the records eg */

$categorySql="SELECT cID,cDescription FROM category WHERE categoryID=$category"; 

$result = mysql_query($categorySql);

/*error check*/
if(!$result){
	echo "There has been an error. ";
	print("errno: " . mysql_errno());
	print ("error: " . mysql_error());
}

/* no error so loop through the records*/

echo "<table>";
while ($myrow = mysql_fetch_array($result)) {
	$cID=$myrow["cID"];
	$categoryDescription=$myrow["cDescription"];

	/*then build the table*/

	echo "	<tr>\n";
	echo "	<td bgcolor=\"#ffffff\">$categoryDescription</td>\n";
	echo "	</tr>\n";
}
echo "</table>";
}else{
	/*there is no category variable been passed to the page, so you need to show the search form*/

/*insert existing form code here*/
}
//bottom og html page stuff here

One other point - in your existing code, it appears that your giving the details eg 11 Mile Truck Frame/Axle as the option value - I'd suggest you switch that to the ID for that field as you may otherwise have a problem getting a correct match...

lemme know if this helps...

imbizsol

Well, okay, it's obvious what I'm trying to do now is way over my head. :-) Sorry to keep bothering you with this. I'm going to figure out how to do this even if it kills me.

Where it says insert existing form code here, I should have:

/*insert existing form code here*/
echo "

<form method=\"POST\" action=\"testnewx.php\">
<div align=\"center\"><center><p>Search Categories
<select name=\"category\">
<?php
$query = "SELECT category FROM members ORDER BY category";
$result = mysql_query($query, $connection);
if(mysql_num_rows($result)) {
// we have at least one user, so show all users as options in select form
while($row = mysql_fetch_row($result))
{
print("<option value=\"$row[0]\">$row[0]</option>");
}
} else {
print("<option value=\"\">No records were found, sorry.</option>");
}
?>
</select><input type=\"submit\" value=\"Submit\"></p></center></div>
</form>
<form method=\"POST\" action=\"testnewx.php\">
<div align=\"center\"><center><p>Search Alphabetically
<select name=\"company\">
<?php
$query ="SELECT company FROM members ORDER BY company";
$result = mysql_query($query, $connection);
if(mysql_num_rows($result)) {
// we have at least one user, so show all users as options in select form
while($row = mysql_fetch_row($result))
{
print("<option value=\"$row[0]\">$row[0]</option>");
}
} else {
print("<option value=\"\">No records were found, sorry</option>");
}
?>
</select><input type=\"submit\" value=\"Submit\"></p></center></div>
</form>
</body>
</html>
";
}
?>

However, when I keep all that in there, I'm getting t-string parse errors on the lines that have double quotes on them (ie., a t-string error on the line: $query ="SELECT company FROM members ORDER BY company"; and on the print lines too).

I went and changed the double quotes to single quotes and I didn't get any errors then, but then I didn't get any categories or companies showing up either. All I got was the else print options of "no records were found, sorry".

http://www.madisonheightschamber.org/testnewx.php

Why would I be getting t-string errors on those particular lines and do I need to change them somehow?

By the way, thanks so much for trying to help me get this code working. I really appreciate it. :-)

achuka

This seems to be be a very similar problem to the one I'm trying to resolve...

except I think I'm one step ahead, in the seanse that I have successfuuly jumped from the search options page to a a list of the reults...

I can share the code (for a book-based site) if it helps - my next problem, which I HAVEN'T resolved yet, is to LIMIT the returns to,say, 20 results and then include NEXT / PREVOUS links as appropriate - that's getting a little tricky, but this pm I bought Larry Ullman's PHP Advanced (Peachpit Press) which has a chapter (3) on that, so I'm hoping to get things sorted over this weekend (a 3-day holiday in the UK!)

Michael

Michael Thorn
ACHUKA Children's Books UK
http://www.achuka.co.uk/index2.html
info@achuka.co.uk

TodManPlaa

OK - you'll get there eventually!

I took a look at the code you posted and there looked to be a few problems with it - you were backslashing quotes were it wasn't necessary and you had some nested php delimiters - to cut a long story short, I've run through and corrected it (bear in mind its 7am Sunday here, so I THINK I've corrected it all 😃 )

so here we go:

<form method="POST" action="testnewx.php">
<div align="center"><center><p>Search Categories
<select name="category">
<?php
$query = "SELECT category FROM members ORDER BY category"; 
$result = mysql_query($query, $connection); 
if(mysql_num_rows($result)) { 
	// we have at least one user, so show all users as options in select form 
	while($row = mysql_fetch_row($result)) { 
		print("<option value=\"$row[0]\">$row[0]</option>"); 
	} 
} else { 
	print("<option value=\"\">No records were found, sorry.</option>"); 
} 
?> 
</select><input type="submit" value="Submit"></p></center></div> 
</form> 
<form method="POST" action="testnewx.php"> 
<div align="center"><center><p>Search Alphabetically 
<select name="company"> 
<?php 
$query ="SELECT company FROM members ORDER BY company"; 
$result = mysql_query($query, $connection); 
if(mysql_num_rows($result)) { 
	// we have at least one user, so show all users as options in select form 
	while($row = mysql_fetch_row($result)) { 
		print("<option value=\"$row[0]\">$row[0]</option>"); 
	} 
} else { 
	print("<option value=\"\">No records were found, sorry</option>"); 
} 
?> 
</select><input type="submit" value="Submit"></p></center></div> 
</form> 
</body> 
</html>

Note if you are pasting that into the other code - remember to close and open the PHP tags before and after it - as the start and finish of this is plain old HTML - no parsing required.

so try pasting that in and lemme know how you go - I lloked at the file you suggested and you've got obvious problems with that, as it appears that you have PHP code in there that isn't bracketed to <?php tags, so it hasn't been parsed.

If the above still doesn't work, then remove the second search function, and we'll work just on the one - get that one to work, then it should be simple to get the other one to work.

I'll be around all morning - lemme know how you go.

TodManPlaa

Hey again,

For Achuka's problem - look at using LIMIT

For example:
SELECT fish,chilli,basil FROM food WHERE fish="tasty" ORDER BY chilli LIMIT 10;

Will select the top 10 records where the fish is tasty and the spiciest

SELECT fish,chilli,basil FROM food WHERE fish="tasty" ORDER BY chilli LIMIT 11,10;

Will select the next 10 records where the fish is tasty and the spiciest

and so on.

Format for LIMIT is LIMIT m,n where m is the starting row in the set and n is the number of rows returned. So if you pass m in the querystring (or however), you should be set.

cheers

imbizsol

Okay, I think I'm starting to get somewhere; however, I'm getting a syntax error now that reads:

There has been an error. errno: 1064error: You have an error in your SQL syntax near 'Services' at line 1

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mhcham/public_html/testnewx.php on line 34

Line 34 of my code is the while statement.

Any ideas on what to look for to fix this error?

http://www.madisonheightschamber.org/testnewx.php

For achuka: I did an alphabetical search script for this same script I'm working on and I figured out how to limit the records to a page and have next/previous links. It's actually one of the things I didn't have much problem doing, but I see that the post above pretty much explains it :-)

Thanks again,

TodManPlaa

OK, lets get more details on the error - after each of the two lines that has the SELECT statement, enter the following:

echo $query;
/*where $query is the variable that you're setting the SQL query to*/

As this will display the SQL statement to the screen before the error hangs it, so then we'll have a better idea of what is causing the error. My guess at the moment is that it is quote related - be sure that your statement is along the lines of:
SELECT a,b,c FROM d where a='something';
(note the quotes)

NOTE: this must go after the SELECT statement, but before the line that runs the query to the db - something like

$result = mysql_query($query, $connection);

This is one of the main reasons you have error checking - otherwise often you can go for ages without even knowing that a script isn't doing what it should be!

see how you go
/s

imbizsol

Woohoo, I fixed the error and touched up my code and now it's working. Thanks sooooo much. However, I do have one question.

Is there a way to have the category title show up only once on a results page? If you go to http://www.madisonheightschamber.org/textnewx.php and click on Automotive Services, you should have two records pull up, except they both have the category title of Automotive Services above their records. I want the category title to show up only once at the top of the page, not for every record.

Hope that makes sense.

TodManPlaa

Glad to hear you got it working - its been a welcome distraction from the work I should have been doing!!

OK with your question, you need to look at some more SQL commands to to narrow down your search - particularly GROUP BY and DISTINCT.

As I'm not sure how your table is structured it's a bit tricky to explain just what to do. But I'm sure if you look at some of the other posts in this forum you'll find the answer you're looking for.

Glad to have been of help

Cheers

imbizsol

Thanks for the help. I'll take a look around and see what I can find :-)

For Achuka, here's the code snippet I used to limit the rows of data displayed on a page and to add next/previous links:

<?php
// Connect to the database
$db = mysql_connect("localhost","xxxxx","xxxxx");
mysql_select_db (members);
// Ask the database for the information from the table
// Show 5 rows of data, starting from whatever position $start is equal to
if (!isset($start)) $start = 0;
$query = "SELECT * FROM members ORDER BY company LIMIT " . $start . ", 5";
$result = mysql_query($query);
// Now we print out the results, starting by making a table
echo ("<table border = '0' width = '70%' align='center'>");
echo "<font color=navy><u>Restaurants</u></font><p>\n";

while ($rows = mysql_fetch_row($result))
{
printf("<tr><font color=navy>%s<br>%s<br>%s<br>%s, %s %s<br>Phone: %s<br>Fax: %s<br>Email: %s<br>Website: %s<p></font></tr>\n", $rows[2], $rows[3], $rows[4], $rows[5], $rows[6], $rows[7], $rows[8], $rows[9], $rows[10], $rows[11]);
}
// do another query to get the number of rows in our table
$query = "SELECT count(*) as count FROM members";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
$numrows = $row['count'];
if($start > 0)
echo "<a href=\"" . $PHP_SELF . "?start=" . ($start - 5) .
"\">Previous</a><BR>\n";
if($numrows > ($start + 5))
echo "<a href=\"" . $PHP_SELF . "?start=" . ($start + 5) .
"\">Next</a><BR>\n";
// Finally we close off the table
echo "</table>";
?>

Hope this helps.

achuka

I adapted this for my own database and it worked fine
On The First Page
but the all the Next Pages provide blank tables when there should be lots of results....

Is it something to do with needing to recapture the search criteria (from a search form page) each time, and if so what do I need to add?
Michael info@achuka.co.uk

<?php
// Connect to the database
$db = mysql_connect("localhost","username","");
mysql_select_db (nonfiction);
// Ask the database for the information from the table
// Show 5 rows of data, starting from whatever position $start is equal to
if (!isset($start)) $start = 0;
$query = "SELECT * FROM poetry WHERE ".$searchtype." like '%".$searchterm."%' order by ".$orderby." LIMIT " . $start . ", 5";
$result = mysql_query($query);
// Now we print out the results, starting by making a table
// The following HTML only for testing purposes...
//Print a table with the following column headings
echo '<table align="center" cellpadding="2" cellspacing="2" border="1">
<tr align="center">
<td align="center"><b>Title</b></td>
<td align="center"><b>Name</b></td>
<td align="center"><b>Month</b></td>
<td align="center"><b>Year</b></td>
<td align="center"><b>Publisher</b></td>
<td align="center"><b>Rating</b></td>
<td align="center"><b>Comment</b></td>
</tr>';

//Print each item.
$query_result=@($query);
while ($row=@mysql_fetch_array($query_result)) {

	echo " <tr align=\"center\">
	<td align=\"left\">$row[Title]</td>
	<td align=\"left\">$row[Name]</td>
	<td align=\"left\">$row[Month]</td>
	<td align=\"left\">$row[Year]</td>
	<td align=\"left\">$row[Publisher]</td>
	<td align=\"left\">$row[Rating]</td>
	<td align=\"left\">$row[Comment]</td>
	</tr>";

}

// do another query to get the number of rows in our table
$query = "SELECT count(*) as count FROM nonfiction";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
$numrows = $row['count'];
if($start > 0)
echo "<a href=\"" . $PHP_SELF . "?start=" . ($start - 5) .
"\">Previous</a><BR>\n";
if($numrows > ($start + 5))
echo "<a href=\"" . $PHP_SELF . "?start=" . ($start + 5) .
"\">Next</a><BR>\n";
// Finally we close off the table
echo "</table>";
?>

imbizsol

Ack, I don't know what happened. My code for the search form was working fine yesterday, then today I imported my database into MySQL and now when I do anything with the form, it duplicates the categories. For instance, if I have 4 companies under the category of Accountants, I get four categories reading Accountants in the drop down list.

You can see what I mean by going here:

http://www.madisonheightschamber.org/testnewx.php

There are three accountants in the database, and accountants shows up three times in the drop down list.

What could have happened?

Panic time? :-)

achuka

Surely the search form should function
independently of the php results program
or are you generating this page by php too?
Alternatively is this my pwn Big Problem
- that my search form is not par tof the
PHP program - see below....

The source code for the url you posted
gives the Accountant triplicate

Here's my own search form
which I created in HTML to link
with a php page

(the one I'm having difficulty with
regarding the search results on the Next pages)

Choose search type:<br>
<select name="searchtype">
<option value="Name">Name
<option value="Publisher">Publisher
<option value="Year">Year
<option value="Month">Month
<option value="Rating">Rating
</select>
<br>
Enter search term:<br>
<input name="searchterm" type=text>
<br>
<input type=submit value="Search">
</input>
<br>
Order by:<br>
<select name="orderby">
<option value="Year">Year
<option value="Publisher">Publisher
</select>
</form>

imbizsol

TodManPlaa had helped me get my code working properly, and yesterday, it was pulling the categories from my database just fine, but now today, it's pulling each category more than once. I don't have any idea what might have happened. I haven't touched anything in the code.

Here is my entire code:

<?php
// create connection
$connection = mysql_connect("localhost","xxxxx","xxxxx")
or die("Couldn't make connection.");

// select database
$db = mysql_select_db("members", $connection)
or die("Couldn't select database.");

if ($category){
/if this variable is present on the page,
it means that the user has selected a particular category
from the form and hit submit
so run a query to get the records eg /

$categorySql="SELECT ID, category, company, contact, address, city, state, zip, phone, fax, email, url FROM members WHERE category='$category'";

echo $categorySQL;
/where $query is the variable that you're setting the SQL query to/

$result = mysql_query($categorySql);

/*error check*/
if(!$result){
    echo "There has been an error. ";
    print("errno: " . mysql_errno());
    print ("error: " . mysql_error());
}

/* no error so loop through the records*/

echo "<table width=80% align=center>";
while ($myrow = mysql_fetch_array($result)) {
    $ID=$myrow["ID"];
    $category=$myrow["category"];
    $company=$myrow["company"];
    $contact=$myrow["contact"];
$address=$myrow["address"];
    $city=$myrow["city"];
    $state=$myrow["state"];
    $zip=$myrow["zip"];
    $phone=$myrow["phone"];
$fax=$myrow["fax"];
$email=$myrow["email"];
    $url=$myrow["url"];

    /*then build the table*/

    echo "    <tr>\n";
    echo "    <td bgcolor=\"#ffffff\"><u><b>$category</b></u><p> $company<br>$contact<br>$address<br>$city, $state $zip<br>Phone: $phone<br>Fax: $fax<br> $email<br>$url<p></td>\n";
    echo "    </tr>\n";
}
echo "</table>";

}else{
/there is no category variable been passed to the page, so you need to show the search form/

/*insert existing form code here*/

}
?>
<form method="POST" action="testnewx.php">
<div align="center"><center><p>Search Categories
<select name="category">
<?php
$query = "SELECT category FROM members ORDER BY category";
echo $query;
/where $query is the variable that you're setting the SQL query to/

$result = mysql_query($query, $connection);
if(mysql_num_rows($result)) {
// we have at least one user, so show all users as options in select form
while($row = mysql_fetch_row($result)) {
print("<option value=\"$row[0]\">$row[0]</option>");
}
} else {
print("<option value=\"\">No records were found, sorry.</option>");
}
?>
</select><input type="submit" value="Submit"></p></center></div>
</form>

The search form is working fine now except for the fact that it's listing each category that has more than one record more than once in the search form now.