jQuery 选择选项 - CodeIgniter

Question

使用codeigniter生成表单。表单有一个 form_dropdown()

  <div class="control-group">
      <?php 
        foreach ($clients as $client => $clientValue) {
           $option[$clientValue['clientName']] = $clientValue['clientName'];
        }
        echo form_label('Client Name:', 'client');
        echo form_dropdown('client', $option);
      ?>
</div><!-- /control-group -->

输出

<select name="client">
<option value="Mike">Mike </option>
<option value="Sissel Bygton">Sissel Bygton</option>
<option value="Calli Crass">Calli Crass</option>
</select>

Javascript/jQuery

var clientName = json.selected_gallery[0].clientName; //clientName retrieved from DB

//alert box to confirm clientName for this example lets say it's Mike
alert(clientName)//output Mike

//Client Select name
$('select.client').val(clientName);

根据我通过其他问答收集的信息，我认为我的 jQuery 可以解决问题。我遗漏了什么吗？

Answer 1

<select name="client">

选择器应该是

$('select[name="client"]').val(clientName);

如果您想使用类访问它，那么

<select class="client">

上面的选择器是

$('select.client').val(clientName);