pull data by drop down menu

ychan623

Hello, I'm new to php, I'm working on a website for employment. I want user to be able to pull data from mysql by using selection from drop down menu. I have a table call jobs, inside this table there are 5 columns: JobType, Place, Wage, Phone, Date. I created the drop down menu and the php filter code (not sure if i'm coding it right). But went I run it on IE, the page just refresh. It didn't echo anything on the page. Here is my code...... Please Help!!!

This is in the selection page:

<form action="selectform.php" method="POST">
<select name="jobselect">
<option value="" selected="selected">All Jobs</option>
<option value="Cook1">1st Cook</option> 
<option value="Cook2">2nd Cook</option> 
<option value="Helper">Kitchen Helper</option> 
<option value="Bagger">Bagger</option> 
<option value="DishWasher">Dish Washer</option> 
<option value="Hostress">Hostress</option> 
<option value="Waiter">Waiter</option> 
<option value="Waitress">Waitress</option> 
<option value="BusBoy">Bus Boy</option> 
<option value="Fryer">Deep Fryer</option> 
<option value="Cashier">Cashier</option> 
<option value="Driver">Driver</option> 
</select> 
<input type="Submit" value="Submit" name="Submit">
</form>

This is in the filtered page:

<?php
mysql_connect ("localhost","root") or die (mysql_error());            mysql_select_db ("nindu") or die (mysql_error());

?>
<?php

$Job = isset($_POST["jobselect"]);
$Query = "SELECT * FROM jobs WHERE JobType='$Job'";

//RUN THE QUERY
	$result = mysql_query($Query) or die (mysql_error());
while ($row = mysql_fetch_array($result)) {
	echo "<br />";
	echo $row['JobType'] . "-" . $row['Place'] . "-" .
                         $row['Wage'] . "-" . $row['Phone'] . "-" . 
                         $row['Date'];
	echo "<br />";
}
?>

djjjozsi

Its wrong:

	$Job = isset($_POST["jobselect"]);
	$Query = "SELECT * FROM jobs WHERE JobType='$Job'";

try:

<?php
if (isset($_POST["jobselect"])) {
    $Job = mysql_real_escape_string($_POST["jobselect"]);
    $Query = "SELECT * FROM jobs WHERE JobType='$Job'";

/* rest of you code!  */
} 
?>

You should use mysql_fetch_assoc instead when you fetch the results, because mysql_fetch_array gives you numerical indexes too, if you don't use it you just waste the memory.

ychan623

Thanks Djjjozsi, Finally it worked. But I need another code to pull all from this table, since All Jobs are preselected. So when a user submitted it with All Jobs, he will see all available jobs.
Thanks again, You're the man 🙂

bradgrafelman

Take the WHERE clause out of the query, and only append it to the query string if $POST['jobselect'] isn't empty, e.g.:

if(!empty($_POST['jobselect']))
    $query .= ' (WHERE condition)';

ychan623

Thanks bradgrafelman. when i left out the where clause all the job will be displayed. than use cannot filter by job type. or am i not doing it right.

djjjozsi

ychan623;10948433 wrote:
Thanks bradgrafelman. when i left out the where clause all the job will be displayed. than use cannot filter by job type. or am i not doing it right.

You should ask with source code.

ychan623

I still need help with my website. I want user to be able to search all jobs if they submit search with All Jobs selected. code below

// THIS IS IN THE SELECTION PAGE
<form action="selectform.php" method="post">
<select name="jobselect">
<option value="All" selected="selected">All Jobs</option>
<option value="Cook1">Cook1</option>
<option value="Cook2">Cook2</option>
<option value="Helper">Helper</option>
<option value="Bagger">Bagger</option>
<option value="DishWasher">DishWasher</option>
<option value="Hostress">Hostress</option>
<option value="Waiter">Waiter</option>
<option value="Waitress">Waitress</option>
<option value="BusBoy">BusBoy</option>
<option value="Fryer">Fryer</option>
<option value="Cashier">Cashier</option>
<option value="Driver">Driver</option>
</select>
<input type="Submit" value="Submit" name="Submit">
</form>

// THIS IS IN THE FILTERED PAGE
<?php
mysql_connect ("localhost","root") or die (mysql_error());

mysql_select_db ("nindu") or die (mysql_error());

?>
<?php

if (isset($_POST["jobselect"])) { 
$Job =  mysql_real_escape_string ($_POST["jobselect"]); 
$Query = "SELECT * FROM jobs WHERE JobType='$Job'";

//RUN THE QUERY
$result = mysql_query($Query) or die (mysql_error());
while ($row = mysql_fetch_array($result)) {
echo "<br />";
echo $row['JobType'] . " - " . $row['Place'] . " - " . $row['Wage'] . " - " .
$row['Phone'] . " - " . $row['Date'];
echo "<br />";
}
}

// Everything above this part works ok

<?php
//but I also want php to pull all jobs from the table if user submit with All Jobs selected.
//Please correct code if wrong
$Query2 = "SELECT * FROM jobs" ;
if ($Job == "All Jobs") {
echo $Query2 ;

	$result = mysql_query($Query2) or die (mysql_error());
while ($row = mysql_fetch_array($result)) {
echo "<br />";
echo $row['JobType'] . " - " . $row['Place'] . " - " . $row['Wage'] . " - " . 
	 $row['Phone'] . " - " . $row['Date'];
echo "<br />";

}

bradgrafelman

I already posted one way to solve this above. Start with no WHERE clause in the query. Then use an if() statement to see if something other than 'All Jobs' was POST'ed and, if so, add in the appropriate WHERE clause.

ychan623

Hello bradgrafelman thanks for responding. I try your suggestion but I'm sure if I got the code right. Because it return no result. Here is my new code. Could you please correct it if i'm wrong
, Please 🙂

// THIS IS IN THE FILTERED PAGE
<?php
mysql_connect ("localhost","root") or die (mysql_error());

mysql_select_db ("nindu") or die (mysql_error());

?>
<?php

if (isset($_POST["jobselect"])) { 
$Job =  mysql_real_escape_string ($_POST["jobselect"]); 
$Query = "SELECT * FROM jobs";
		if(!$job ['All Jobs']){
		echo "SELECT * FROM jobs WHERE JobType='$Job'"; 
	}

ychan623

I'm NOT sure if I got the code right

Weedpacket

When posting PHP code, please use [noparse]

...

[/noparse] around it so that the formatting doesn't fall over (the forum's FAQ describes additional tags for things like HTML and plain code). With those tags (and consistent layout), code will be much more legible and many classes of error become obvious. See your original post for examples.

ychan623

Help, I need another code to pull all from this table, since All Jobs are preselected. So when a user submitted it with All Jobs, he will see all available jobs.

// THIS IS IN THE FILTERED PAGE
<?php
mysql_connect ("localhost","root") or die (mysql_error()); 
mysql_select_db ("nindu") or die (mysql_error());

?>
<?php

if (isset($_POST["jobselect"])) { 
$Job = mysql_real_escape_string ($_POST["jobselect"]); 
$Query = "SELECT * FROM jobs";
if(!$job ['All Jobs']){
echo "SELECT * FROM jobs WHERE JobType='$Job'"; 
}
//RUN THE QUERY
$result = mysql_query($Query) or die (mysql_error());
while ($row = mysql_fetch_array($result)) {
echo "<br />";
echo $row['JobType'] . " - " . $row['Place'] . " - " . $row['Wage'] . " - " . 
$row['Phone'] . " - " . $row['Date'];
echo "<br />";
}
}

?>
[]/php

ychan623

Help, I need another code to pull all from this table, since All Jobs are preselected. So when a user submitted it with All Jobs, he will see all available jobs.

// THIS IS IN THE FILTERED PAGE
<?php
mysql_connect ("localhost","root") or die (mysql_error()); 
mysql_select_db ("nindu") or die (mysql_error());

?>
<?php

if (isset($_POST["jobselect"])) { 
$Job = mysql_real_escape_string ($_POST["jobselect"]); 
$Query = "SELECT * FROM jobs";
if(!$job ['All Jobs']){
echo "SELECT * FROM jobs WHERE JobType='$Job'"; 
}
//RUN THE QUERY
$result = mysql_query($Query) or die (mysql_error());
while ($row = mysql_fetch_array($result)) {
echo "<br />";
echo $row['JobType'] . " - " . $row['Place'] . " - " . $row['Wage'] . " - " . 
$row['Phone'] . " - " . $row['Date'];
echo "<br />";
}
}

?>