Question

I’m stuck with this bit of code. By all accounts it should work, I’ve been staring at it for 2 hours now, I have no idea what is wrong.

I have two selections with options in divs. When one is selected with the right parent id, the sub category should appear, it does not for some reason though I feel I’ve done it all right – here is my html/php for the main page:

<li>
            <label for="Catid">Pick a Category:</label>
            <select id="Catid">
              <?php
                  $Products = New Products();
                  $Products->form_Cat_Picker();
                ?>
            </select>
          </li>
          <li>
            <label for="Subcatid">Sub Category:</label>
              <?php
                  $Products = New Products();
                  $Products->form_Subcat_Picker();
                ?>
          </li>
          <script type="text/javascript">
            $(document).ready(function() {
                $('#Catid').change(function(){
                    var optvalue = $(this).val(),
                    div = $('#' + 'parentid' + optvalue);
                    $('div').hide();
                    div.show();
                });
            });​
          </script>
          <li>

And here are the two functions:

function form_Cat_Picker() {
    $mysqli = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME);
    if (!$mysqli) {
        die('There was a problem connecting to the database.');
    }       
    $catPicker = "SELECT Catid, Catname
            FROM ProductCats
            ORDER BY Catid";
    if ($Result = $mysqli->query($catPicker)){
        if (!$Result) {
            echo 'Could not run query: ' . mysql_error();
            exit;
        }
        echo '<option value="">'.''.'</option>';
        while ($row = $Result->fetch_assoc()) {
            echo '<option value="'.$row["Catid"].'">'.$row["Catname"]."</option>";
            echo '</div>';
        }
    }
    $mysqli->close();
}
function form_Subcat_Picker() {
    $mysqli = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME);
    if (!$mysqli) {
        die('There was a problem connecting to the database.');
    }       
    $catPicker = "SELECT Subcatid, Subcatname, Parentid
            FROM ProductSubCats
            ORDER BY Subcatid";
    if ($Result = $mysqli->query($catPicker)){
        if (!$Result) {
            echo 'Could not run query: ' . mysql_error();
            exit;
        }
        $counter = 0;
        while ($row = $Result->fetch_assoc()) { 
            if ($counter >= 1) {
                if ($last_id != $row['Parentid']){
                    echo '</select>';
                    echo '</div>';
                }
            }
            if ($last_id != $row['Parentid']){
                    echo '<div id="parentid'.$row['Parentid'].'" style="display:none">';
                    echo '<select id="Subcatid">';
                }
            echo '<option value="'.$row["Subcatid"].'">'.$row["Subcatname"]."</option>";
            $last_id = $row['Parentid'];
            $counter++;
        }
        echo '</select>';
        echo '</div>';
    }
    $mysqli->close();
}

These output the following:

<li>
<label for="Catid">Pick a Category:</label>
<select id="Catid">
<option value=""/>
<option value="1">Sample 1</option>
<option value="2">Sample 2</option>
<option value="3">Sample 3</option>
<option value="4">Sample 4</option>
<option value="5">Sample 5</option>
<option value="6">Sample 6</option>
<option value="7">Sample 7</option>
</select>
</li>
<li>
<label for="Subcatid">Sub Category:</label>
<div id="parentid1" style="display: none;">
<select id="Subcatid">
<option value="1">Sub Sample 1</option>
<option value="2">Sub Sample 2</option>
</select>
</div>
</li>

So the output looks fine. The jquery seems correct.
But when I try to run it – the div id parentid1 does not appear when Sample 1 is selected.

Am I going blind or have I done something wrong? Thanks

EDIT

even works in jfiddle:
http://jsfiddle.net/VGDwm/
And I called in the jquery library properly with:

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js" type="text/javascript"/>

What could be wrong? Should the jquery be after or before the divs? Or is this an issue with calling in from php?

Answer 1

1. Don’t re-connect to the database every function
2. Re-type the last line of the jquery, it has a hidden character at the end.

so the line highlighted here:

      <script type="text/javascript">
        $(document).ready(function() {
            $('#Catid').change(function(){
                var optvalue = $(this).val(),
                div = $('#' + 'parentid' + optvalue);
                $('div').hide();
                div.show();
            });
        });​    <----- THIS LINE ------>
      </script>