CODEWITHSUNDEEP
phpopencart2.x
Sepika
Sepika

Reputation: 13

Multiple options in select in php form

I have this simple code in my product_form.tpl php file of opencart 2.3:

<select name="type" id="type" class="form-control">
<?php if ($type) { ?>
<option value="1" selected="selected">Flower</option>
<option value="0">Plant</option>
<?php } else { ?>
<option value="1">Flower</option>
<option value="0" selected="selected">Plant</option>
<?php } ?>
</select>

Everything fine and simple, but i need at least 1 more option for this select to add:

<option value="2">Bouquet</option>

Could you please tell me how this should look like in case of 3 options... Thanks in advance!

Upvotes: 1

Views: 199

Answers (4)

Darker
Darker

Reputation: 83

If you really want to go this way, then:

<select name="type" id="type" class="form-control">
<?php if ((int)$type === 0) { ?>
<option value="2">Bouquet</option>
<option value="1">Flower</option>
<option value="0" selected="selected">Plant</option>
<?php } elseif ((int)$type === 1) { ?>
<option value="2">Bouquet</option>
<option value="1" selected="selected">Flower</option>
<option value="0">Plant</option>
<?php } elseif ((int)$type === 2) { ?>
<option value="2" selected="selected">Bouquet</option>
<option value="1">Flower</option>
<option value="0">Plant</option>
<?php } ?>
</select>

But I would suggest using the way Jelmer Jellema suggested.

Upvotes: -1

Jelmer Jellema
Jelmer Jellema

Reputation: 1116

Sometimes it's easier to just write some structured code..

<?php
 $options = array(
        "1" => "flower",
        "0" => "Plant",
        "2" => "Bouquet",
        "3" => "Rock"
 );
 $currentval = 2;
?>

<select name="type" id="type" class="form-control">
   <?php
   foreach($options as $val => $label):
    ?>
      <option value="<?=$val;?>" <?=$currentval == $val ? 'selected' : ''?>><?=$label;?></option>
   <?php
    endforeach;
   ?>

Upvotes: 1

pavel
pavel

Reputation: 27082

You chose a bad approach as I wrote in comment under question. 

<select name="type" id="type" class="form-control">
    <option value="1" <?php echo isset($type) && $type == 1 ? 'selected' : '' ?>>Flower</option>
    <option value="0" <?php echo isset($type) && $type == 0 ? 'selected' : '' ?>>Plant</option>
    <option value="2" <?php echo isset($type) && $type == 2 ? 'selected' : '' ?>>Bouquet</option>
</select>

Another way is using array and write the dropdown using foreach loop.

isset($type) can be taken above the code, you need to check if variable exists to prevent notice of undefined variable.

There can be something like

$type = isset($type) ? $type : NULL; // or any default value like 0

In options then will be just <?php echo $type == 1 ? 'selected' : ''; ?>.

Upvotes: 2

Jasson Rojas
Jasson Rojas

Reputation: 299

You can use a siwtch case block instead of bunch of if's you can check the documentation here https://www.w3schools.com/PhP/php_switch.asp and you can compare the value of $type more organized ! Good Luck !

Upvotes: -1

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