populate dropdown list from database when editing form MySQL

Question

I am pretty new to PHP and am having a question about dropdown lists. I am trying to get the list to pull from the database and populate the value when a user edits a form but it is not currently working. There are a few examples of this exact same thing on here but I can't quite get it working, it is likely a syntax error on my end...

Here is my code:

echo '<p><label>Is this project targeted toward?</label><select name="proj_targ_tow"><option value="Select...">Select...</option><option value="National Site">National Site</option><option="Local Site">Local Site</option><option value="Regional Site">Regional Site</option><option value="Other">Other</option></select></p>';

And here is the logic to populate the value from the database, the row I am trying to pull from is 'proj_targ_tow'...

    $typesArray = array ( 'Select..', 'National Site', 'Local Site', 'Regional Site', 'Other' );
$selectedType = '';
echo 'as;ldfjas;lfmawoiealknfsliu2047a   ' . $row['proj_targ_tow'] . '<br />';
foreach($typesArray as $value){
    if($value == $row['proj_targ_tow']) {
        $selectedType = 'selected="selected"';
} 
echo '<option value="' . $value . '" ' . $selectedType . '>' . $value . '</option>';
}

Can any of you coding gods out there help me out?

Also, I'm not sure if you intended it, but make sure you do: $selectedType = '' inside the foreach loop (at the start). — Tushar
– Tushar, Commented May 4, 2012 at 14:55
@TusharDhoot - when I echo it out it returns the value from the database (National Site)... — user1288091
– user1288091, Commented May 4, 2012 at 14:55

Daniel Bidulock · Accepted Answer · 2012-05-04 15:01:33Z

2

It looks to me like the $value in your echo statement is out of scope... what happens when you do this?

$typesArray = array ( 'Select..', 'National Site', 'Local Site', 'Regional Site', 'Other' );

foreach($typesArray as $value){
    $selectedType = '';

    if($value == $row['proj_targ_tow'])
        $selectedType = 'selected="selected"';

    echo '<option value="' . $value . '" ' . $selectedType . '>' . $value . '</option>';
}

edited May 4, 2012 at 15:01

answered May 4, 2012 at 14:55

Daniel Bidulock

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2 Comments

user1288091 Over a year ago

this is pretty much it, except for I needed an else to populate all the other values that are NOT selected. Thanks for your help!!!

Daniel Bidulock Over a year ago

I'm happy to have been of assistance.

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