Can't seem to capture a variable in a chained select

[86Stang]

I'm *this* close to having a chained select running but for some reason it doesn't seem to be picking up a variable.

 

<?php

require ('inc/connection.php');

//seeming that we are just submitting and refreshing to the one page we need to check if the post variable is set, and if so a couple of other variables are set
if(!isset($_POST['state'])) {
    $next_dropdown = 0;
}
else {
    //When set this variable reveals the next drop down menu
    $next_dropdown = 1;
    //this variable keeps the previous selection selected
    $selected = $_POST['state'];
}
?>

<form name="form" method="post" action="">
    <select name="state" style="font-size:20px;">
        <option value="NULL">State</option>
        <?php
        $query = "SELECT id, name FROM state ORDER BY name ASC";
        $result = mysql_query($query);        
        while($row = mysql_fetch_array($result))
        {?>
		<option value="<?php echo $row[0]; ?>" onClick="document.form.submit()" <?php if(isset($selected) && $row[0] == $selected) {echo "selected='selected'";} ?>><?php echo $row[1]; ?></option>\n";
        <?php }
        echo '</form>\n';        
        //this is where the other form will appear if the previous form is submitted
        if($next_dropdown == 1) {?>               
            <form name="form2" action="" method="post">
            <select name="city">
            <option value="NULL">City</option>
            <?php
	    $query2 = "SELECT * FROM city WHERE state_id = " . $row[0];
            $result2 = mysql_query($query2);
            while($row2 = mysql_fetch_array($result2))
            { ?>
            	<option value="<?php echo $row2[0]; ?>" onClick="document.form2.submit()"><?php echo $row2[1]; ?></option>
            <?php }?>
            </select>
            </form>
        <?php }
?>

 

The state drop down works fine.  Once a state is selected, it will display the city drop down.  However, the city drop down never populates.  It's as though it forgets what $row[0] is.  Any thoughts?

[Gayner]

SELECT * FROM city WHERE state_id = " . $row[0];

were is the . "??

 

Are u sure it's even gathering the mysql data from that query? i would put a error function in it to see.

[86Stang]

I don't need the . " but I added it anyway and it doesn't change anything.

 

No errors but it doesn't seem to be gathering the data.  Any thoughts on why?

[Gayner]

SELECT * FROM city WHERE state_id = " . $row[0];

were is the . "??

 

Are u sure it's even gathering the mysql data from that query? i would put a error function in it to see.

 

For now i dont see why it's not catching the data and populating the list, just that query looks weird w/o proper '. .' or ". ."

 

 

 

Try just taking out the ' and . and just use the normal variable .. i mean iuno just guess and check bro.. looks right.. sorry maybe other fortunate helpers might help u.

 

[Psycho]

Keep your code organized and it is much easier to work with and debug. This is not tested

 

<?php

require ('inc/connection.php');

//Get selected state if set
$selectedState = (isset($_POST['state'])) ? $_POST['state'] : false;

//Create the State options
$stateOptions = "<option value=\"\">State</option>\n";
$query = "SELECT id, name FROM state ORDER BY name ASC";
$result = mysql_query($query);        
while($row = mysql_fetch_array($result))
{
    $selected = ($selectedState && $row[0]==$selectedState) ? ' selected="selected"' : '';
    $stateOptions .= "<option value=\"{$row[0]}\" onClick=\"this.form.submit();\"{$selected}>{$row[1]}</option>\n";
}

//Create the City options
if ($selectedState===false)
{
    $cityOptions = "<option value=\"\">Select a state</option>\n";
}
else
{
    $cityOptions = "<option value=\"\">Select a state</option>\n";
    $query = "SELECT * FROM city WHERE state_id = {$selectedState}";
    $result = mysql_query($query);
    while($row = mysql_fetch_array($result))
    {
        $stateOptions .= "<option value=\"{$row[0]}\" onClick=\"this.form.submit();\">{$row[1]}</option>\n";
    }
}

?>

<form name="form" method="post" action="">
    <select name="state" style="font-size:20px;">
        <?php echo $stateOptions; ?>
    </select>
</form>

<form name="form2" action="" method="post">
    <select name="city">
        <?php echo $cityOptions; ?>
    </select>
</form>

[86Stang]

Thanks for the rewrite.  It's kicking this error:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /path/to/file.php on line 28

 

Which is here:

 

    while($row = mysql_fetch_array($result))
    {
        $stateOptions .= "<option value=\"{$row[0]}\" onClick=\"this.form.submit();\">{$row[1]}</option>\n";
    }

[Psycho]

Then the query is failing. As my signature states I don't always test my code - especially when the code requires someone else's database.

 

On the line above, change this

    $result = mysql_query($query);

To this:

    $result = mysql_query($query) or die("$query <br><br>".mysql_error());

That should identify the problem.

 

Also, to prevent sql injection, change this

$selectedState = (isset($_POST['state'])) ? $_POST['state'] : false;

To This

$selectedState = (isset($_POST['state'])) ? mysql_real_escape_string($_POST['state']) : false;