CODEWITHSUNDEEP
javascriptphparrays
RNK
RNK

Reputation: 5792

ReferenceError: array is not defined

I have loop in PHP with javascript: First array is printing on console output. But, in second element of array I am getting this error:

ReferenceError: array is not defined

<?php foreach($job_requirements_names as $jrn){ ?>
    <div class="col-md-12 form-group">
            <label class="control-label label-top" for="requirement_<?=strtolower($jrn['name'])?>"><?=$jrn['name']?></label>
            <?php
               $query = new QUERY(array('TABLE'=>$table_name, 'KEY'=>array('name'=>$jrn['name']), 'ASC'=>'n_option'));
               $options = $query->fetchAll();
               unset($query);
               $id = "requirement_".strtolower($jrn['name']);
            ?>
            <script>
               // For countries:
               var data = '<?php echo json_encode($options);?>';
               console.log(data);
               data = JSON.parse(data);
               data = data.map(function (v) { return {id: v.id, text: v.n_option}; });

               $("#<?=$id?>").select2({
                  multiple: true,
                  data: data
               }).select2('data', array());
            </script>

            <input id='requirement_<?=strtolower($jrn['name'])?>' name="requirement_<?=strtolower($jrn['name'])?>" class="col-md-12"/>
    </div>
<?php } ?>

As you can see in javascript array is already defined.

Upvotes: 1

Views: 3677

Answers (1)

Zakaria Acharki
Zakaria Acharki

Reputation: 67525

You should to use Array() instead of array() (Not exist) since JS is a case sensitive language, or simply use [], so it will be :

.select2('data', new Array());
//OR
.select2('data', []);

Hope this helps.

Upvotes: 3

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