PHP POST selection display

Question

I have this code and the problem i'm having with it is that once i select an option and save it won't remember the selection next time.but the post has been registered in the database with proper values. is there a syntax error that caught your attention? thanks for your help.

Platform : PHP,SQLite

<tr>
<?php 
//location
?>
<td class='tdt' class='tdt'><?php te("Department");?>:</td>
<td>
<select id='locations' name='locations'>
<option value=''><?php te("Select");?></option>
<?php 
    foreach ($locations  as $key=>$location ) {
        $dbid=$location['id']; 
        $itype=$location['name'];
        $s="";
        if (($locations=="$dbid")) $s=" SELECTED "; 
            echo "    <option $s value='$dbid'>$itype</option>\n";
    }
?>
</select>

</td>
</tr>

<tr>

Wait.. what? You're checking if `$locations` (which is an array you're iterating THROUGH) is the same as `$dbid` which is a key in a subarray of `$locations`? It'd **never** match. `$locations[0]['id']` would **never** be the same as `$locations`? — h2ooooooo, Sep 04 '13 at 11:17
Well for starters you're not checking `$_POST` anywhere in this code, so obviously it doesn't care what the POST is? — h2ooooooo, Sep 04 '13 at 11:19
I'm not sure how to add $_POST into the mix.I've tried different variations.but with the same outcome.Thanks again — Carter, Sep 04 '13 at 11:22
`if ($_POST["locations"] == $dbid) $s = ' selected="SELECTED"';` ? — h2ooooooo, Sep 04 '13 at 11:22
Thanks but that didn't work either.still not returing the selected value.appriciate your help. — Carter, Sep 04 '13 at 11:25

score 0 · Answer 1 · answered Sep 04 '13 at 11:31

0

You may try this (Assuming you have prepared $locations array)

<select id='locations' name='locations'>
    <option value=''><?php te("Select");?></option>
    <?php 
        foreach ($locations  as $key => $location ) {
            $s = $_POST['locations'] == $location['id'] ? 'selected' : '';
            echo "<option $s value='" . $location['id'] . "'>" . $location['name'] . "</option>";
        }
    ?>
</select>

answered Sep 04 '13 at 11:31

The Alpha

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Hi Sheikh thanks for helping me out.does this array seems functional to you? $locations=$r['locations']; – Carter Sep 04 '13 at 11:38
I assumed that, you have an array like `$locations = array('1'=>'loc1', '2'=>'loc2')`. – The Alpha Sep 04 '13 at 11:40
I have a table that consist of all locations called"locations (id,name)".once user make a selection it will register in the employee table under locations as an INT.That INT is tied to the table that i have mention at the begining.Thanks Sheikh – Carter Sep 04 '13 at 11:45
Then you have to pull that data (locations) from db and supply it to the `foreach` loop, is that you are doing now ? – The Alpha Sep 04 '13 at 11:48
yes.Exactly.But i'm not sure how to setup an array for that.this my SQL query $sql="SELECT * FROM locations order by name"; $sth=$dbh->query($sql); $locations=$sth->fetchAll(PDO::FETCH_ASSOC); thanks – Carter Sep 04 '13 at 11:54
It's ok, also check [this](http://stackoverflow.com/questions/1519872/pdo-looping-throug-and-printing-fetchall) if you need to compare your code. – The Alpha Sep 04 '13 at 11:58

PHP POST selection display

1 Answers1