fill dropdown from array

Question

I have the following codes...

$all  = $ent->getAll($fed_code);

 Array ( [NC] => Array (
    [101] => banana,
    [102] => orange,
    [103] => apple,
    )
    )


$select  = $ent->getSelected($fed_code);

Array
(
[101] => banana,
)

what i want is if the $select value is found in $all array, Then fill the dropdown by the value selected. Here is the Code i have At the moment

<?php
    foreach ($all as $orgKey => $list) { 
?>
<tr><td width="5%">
        <h5>Extra Fruits</h5>
    </td>
    <td width="10%">
        <label class="control-label">Fruits</label>
        <select class="input-xlarge" id="input" name="ent[]">   
            <option value="">Select</option>';

            <?   foreach ($list as $key => $value) { 
                    $selected = in_array($select, $key)?'selected="selected"':'';
                    echo $selected;
                ?>

                <option <?=$selected?> value="<?=$key?>"><?=$value?></option>
                <? } ?>
        </select>
    </td>
                        </tr>

But, It seems there is something wrong, it doesn't select any thing. Some one has some idea please?

rccoros · Accepted Answer · 2014-01-31 08:00:34Z

Do you really need the foreach inside? I tried this and it can already print your select with those in $select as selected.

<select class="input-xlarge" id="input" name="ent[]">   
    <? foreach ($all as $orgKey => $list) {
        $selected = in_array($select, $list) ? 'selected="selected"' : '';
    ?>
       <option <?=$selected?> value="<?=$orgKey?>"><?=$list?></option>
    <? } ?>
</select>

I have just realized that I have interchanged the parameter for the in_array. This should do it. I also added the multiple option in the select tag so if ever we have multiple selected it will be shown.

<select multiple class="input-xlarge" id="input" name="ent[]">   
    <? foreach ($all as $orgKey => $list) {
        $selected = in_array($list, $select) ? 'selected="selected"' : '';
    ?>
       <option <?=$selected?> value="<?=$orgKey?>"><?=$list?></option>
    <? } ?>
</select>

If you don't want to use multiple, the last item that will test positive that it is selected will be the one shown as selected in html.

But the $list will echo Array is it? not the content of the array? how do you solve that, then?
my bad I have interchanged the parameters of the in_array function. $list will only have the value and not array. Please see my update.
Not really it is not working, I have to make it in different process with long code and with a bit of trick. But now it is ok. Thanks anyway. I just give you point for your effort.

MjZac · Accepted Answer · 2014-01-31 07:42:59Z

0

What does printing $selected produce? Also to select an option from select tag, isn't the syntax like:

<select>
<option value="volvo">Volvo</option>
<option value="saab">Saab</option>
<option value="vw">VW</option>
<option value="audi" selected>Audi</option>
</select>

instead of what you give as selected="selected"

answered Jan 31, 2014 at 7:42

MjZac

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