Populating A List Box

[Anon]

Sorry for the PHP newbie question:

Can someone give me an example on how to populate the display/values of a list box within a form using PHP? With ASP it was fairly simple using a server object. Basicly, I want to use a SQL query to populate the display value and send value of a drop down list box. Sounds simple enough...

[Anon]

And indeed it is.
I trust you have the php manual in your other browser window? good.

The steps are these:

open a connection to the database.
(do this at the beginning of your script, because everything that comes after it is useless unless this works. Plus, you might want to use this connection for more than one query in this page)

print the html leading up to the point where you want to list the options.
(so print the "<select name=bla>" tag here too)

formulate your query:
$sql="SELECT field1,field2 FROM table";

then you execute your query (as described in the manual)
$results=mysql_query($sql,$database_link);

then you create a loop to fetch all the records in the results:

while ($row=mysql_fetch_array($results,MYSQL_ASSOC))
{
print "<option value=".$row["field1"].">".$row["field2"]."</option>\n";
};

and print the rest of the page.

[Anon]

Thanks! It works! :-)

Now...how can I extend it? I want to set the default selection of the listbox to the displayed record.

[Anon]

You can mark an entry in the list as "selected" by putting "selected" in the option tag:

<option value="bla" selected>bla</option>

So, in your query you must also request something like an ID that you can compare to the "current id".
If they match, print "selected".

while ($row=mysql_fetch_array($results,MYSQL_ASSOC))
{
print "<option value=".$row["field1"];

if ($row["ID"]==$current_id)
{
print " selected ";
};

print ">".$row["field2"]."</option>\n";
};

[Anon]

Just curious, since you mentioned ASP in your first post; how much easier is this to do in PHP vs. ASP?? If you showed the code for each one, which one can you look at and say oh yeah...i know what they're trying to do?? I'm not that familar with ASP. I tried to learn it but the syntax is just crazy. Thanks for some info...

---John Holmes...

[Anon]

It depends. If you are writing against a WinNT/IIS server it is extremely simple as there is a server-side object that exists. You just place the object within the page and define the query (which is a property of the object). My web server uses ChiliSoft ASP so I don't have the convinience. I guess it would be about the same in this case.

[kjnpf]

What if I am populating a drop down menu from values people have submitted into a database and some of these values occur ceveral times? I only want to show the value once and not have it appearing several times.

I will elaborate. I am putting people in contact with each other around the world and they are from cetain countries. I want the drop down list to show the countries only once in the list and not 5 times for USA if there are 5 entries with people from the USA. How do I show the country only once??

I suspect it will be something I put in my SQL query? I already have it ordered by country but want to remove duplicates!!

[DeltaRho2K]

You would have to add a GROUP BY clause to your SQL query...

$sql="SELECT field1,field2 FROM table GROUP BY countryname"; 

[wayang_mbeling]

I have question related to this thread. I have the code below:

<?
$hostname="localhost";
$username="someone";
$password="somepwd";
$database="dbname";

$connection = mysql_connect($hostname, $username, $password) or die(mysql_error()."<b>Unable to Connect to the Server");
mysql_select_db($database,$connection) or die(mysql_error()."<b>Unable to Connect to the Database!</b>");

$mysql_dropdown_query = "select employee_id,first_name,last_name from personnel where is_active=1";
$dropdown_result = mysql_query($mysql_dropdown_query);

echo '<select name="EmpList" size="1">';
echo '<option value=\"\"></option>';
while ($row = mysql_fetch_row($dropdown_result))
{
echo '<option value=".$row[0].">".$row[1]." ".$row[2]."</option>\n';
};

echo '</select>';

mysql_close();

?>

My problem is that instead of listing the data from database, it lists the part between <option></option> which looks more like this:

".$row[1]."".$row[2]"
".$row[1]."".$row[2]"
".$row[1]."".$row[2]"
".$row[1]."".$row[2]"
".$row[1]."".$row[2]"
....

What I want to show in the dropdown list is the concatenate of First Name and Last Name which are in two separate column in Personnel Table. I also want that when someone select one from the list, it will pass a value of the employee_id.

I will really appreciate if someone can show me what's wrong with the code that I have which does not show the result as I want.

I have been trying with all the code I found from the forum. But, the result is still the same 😕

Thank you in advance.

[DeltaRho2K]

My style is a slight bit different than yours, but I have not had any problems with it, so here you go:
I setup a quick test table, and this worked for me (just as I suspected it would 😉)

echo "<option value=\"".$row[0]."\">".$row[1]." ".$row[2]."</option>"; 

[wayang_mbeling]

hey DeltaRho2k...it works like a charm!! Thanks a lot... 😃

I should pay more attention on those quotes and dots... 🆒

Thanks once again...

[DeltaRho2K]

No problem Wayne...

Those dots and slashes will get you everytime. I still catch myself getting it wrong 😉

BTW, if you don't already use a coding program with color syntax highlighting, I highly recommend you get one. (There are plenty of good free ones out there).

It will help you catch those mistakes much easier in the future.

Jacob