jqtransform不改变从阿贾克斯获得的值 [英] jqtransform don't transform the values received from ajax

问题描述

我有这样的Java脚本：

i have this java script:

    $("#judet div.jqTransformSelectWrapper ul li a").click(function(){     
var jud= $("#judetul1").val(); 
$.ajax({
        type: "POST",
        url: "rental/cms/inc/ajax/cities.php",
        data: { 'jud': jud },
        success: function (msg) {

            $("#oras1").html(msg);

        },
        error: function (xhr, err) {
            alert("readyState: " + xhr.readyState + "\nstatus: " + xhr.status);
            alert("responseText: " + xhr.responseText);
        }
    });     
    });

和这个网站：

<div class=" h">
   <span class="block">Orasul</span>
   <div class="select6" id="oras">

   <select name="oras1" id="oras1"   onchange="zone1();sectorul();">

   </select>
  </div>
   <div class="clear"></div>
   </div>
   <div class="clear"></div>

和这个PHP：

public function get_oras($code3) { 
echo "<option selected='selected' value='0'>Alege oras</option>";
        $code='PPLA';
        $code2='PPLA2';
        $sql="SELECT * FROM `locatii` WHERE (`feature_code`=:code OR `feature_code`=:code2) AND `admin1_code`=:code3 ORDER BY `asciiname` ASC";
        $stmt = $this->dbh->prepare($sql);
        $stmt->bindParam(':code', $code, PDO::PARAM_STR, 30);
        $stmt->bindParam(':code2', $code2, PDO::PARAM_STR, 30);
        $stmt->bindParam(':code3', $code3, PDO::PARAM_INT);

        $stmt->execute();
        foreach ($stmt->fetchAll(PDO::FETCH_ASSOC) as $result) 
        {
             $oras[]="<option value='".$result['geonameid']."'>".$result['asciiname']."</option>";

        }
    return $oras;

}

HTML的JUDETUL：

html for judetul:

<div class=" h">
 <span class="block">Judetul</span>
  <div class="select6" id="judet">
  <?php $judetul=$db->get_judet(); ?>
  <select name="judetul1" id="judetul1"  >
  <option selected="selected">---</option>
  <?php foreach ($judetul as $val=>$k) { ?>
  <option value="<?php echo $val; ?>"><?php echo $k; ?></option>
  <?php } ?>
  </select>
  </div>
  <div class="clear"></div>
  </div>
  <div class="clear"></div>

现在的问题是：

The problem is:

没有jqtransform脚本做工非常精细，但如果我有jqtransform不填充选择oras1的。 我认为这是一个问题，因为我变换一种形式，然后再我填充值......这样的值不会转化成我的oras1选择。

without the jqtransform the scripts work very fine, but if I include jqtransform the select oras1 is not populated. I think it's a problem because I transform a form first and then I populate with values...so that values will not be transformed into my oras1 select.

我该如何解决呢？ 我可以改变选择后，我填充它与价值？ 非常感谢！

How can I fix it? Can I transform the select after I populate it with values? Thanks a lot!

推荐答案

你的猜测是正确的，jqTransform转换为文档准备您的选择框。 jqTransform后，您所看到的外观上的选择框是一个包装。 因此，你的Ajax请求后，更新的选择的只是影响了隐藏select元素而不是包装。

Your guess is correct, the jqTransform transform your select box at document-ready. The appearance you see on select box after jqTransform is a wrapper. Therefore, the update in the select's after your ajax request is only affecting the hidden select element but not the wrapper.

您可以尝试重新jqTransfrom的选择框来得到它的权利。

You can try re-jqTransfrom the select box to get it right.

或者，请访问我的博客 http://infrahtml.blogspot.hk/2013/12/jstransform - 选择框，repopulation.html 以更改插件

Or, visit my blog http://infrahtml.blogspot.hk/2013/12/jstransform-select-box-repopulation.html to make changes to the plugins

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