javascript - 我如何通过 jsonencode 数据使用 Javascript 选择下拉值

Question

在我的 html 表单中，我有 9 个下拉值，如果用户操作是“编辑”，那么它将从数据库中获取值并以 jsonencode 格式返回。如下所示，

JSON DATA

 [{"ed_gender":"Male","ed_blood_group":"A-","ed_marital_status":"Single","ed_branch_id":"11","ed_desig_id":"1","ed_job_type":"Permanent","ed_pay_mode":"Cheque"}] 

HTML

 <select name="ed_gender" class="form-control">
    <option value="">Select</option>
    <option value="Male">Male</option>
    <option value="Female">Female</option>
 </select>

 <select name="ed_marital_status" class="form-control">
    <option value="">Select</option>
    <option value="Single">Single</option>
    <option value="Married">Married</option>
 </select> 

我使用 php 尝试了几行代码，它实际上可以工作，但我正在尝试使用 javascript。

PHP

 <select name="ed_marital_status" class="form-control">
    <option value="">Select</option>
    <option <?php if($ed_marital_status=="Single") echo 'selected="selected"'; ?> value="Single">Single</option>
    <option <?php if($ed_marital_status=="Married") echo 'selected="selected"'; ?> value="Married">Married</option>
 </select> 

所以在这里我必须提取 json 值并使下拉值在页面加载时具有“选定”。

编辑:

JSON DATA

 [{"ed_branch_id":"11","ed_desig_id":"1"}] 

HTML

 <select name="ed_job_location" class="form-control">
     <option value="">Select</option>
      <?php
       foreach($get_branches as $branches){
              $branches_id   = $branches->b_id;
              $branches_name = $branches->b_name; 
              $branches_code = $branches->b_code; 
              echo "<option value='$branches_id||$branches_name||$branches_code'>$branches_name</option>";
       }?>
  </select>

  <select name="ed_desig_id" class="form-control">
      <option value="">Select</option>
      <?php
      foreach($get_designation as $designations){
             $designations_id   = $designations->d_id;
             $designations_name = $designations->d_designation; 
             $designations_code = $designations->d_code; 
             echo "<option value='$designations_id||$designations_code'>$designations_name</option>";
       }?>
   </select>

从上面的 json 中，我只得到分支 id 和设计 id，但在这里我有 || 的值。选择选项中的符号，因此我需要找到特定的 id 并将其显示在下拉列表中。

Answer 1

迭代 json 对象数组，并通过名称选择来填充每个元素值:

var json = [{"ed_gender":"Male","ed_blood_group":"A-","ed_marital_status":"Single","ed_branch_id":"11","ed_desig_id":"1","ed_job_type":"Permanent","ed_pay_mode":"Cheque"}];

$(document).ready(function(){
   $.each(json[0], function(index, element) {

       $("[name="+index + "]").val(element);
   });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select name="ed_gender" class="form-control">
    <option value="">Select</option>
    <option value="Male">Male</option>
    <option value="Female">Female</option>
 </select>

 <select name="ed_marital_status" class="form-control">
    <option value="">Select</option>
    <option value="Single">Single</option>
    <option value="Married">Married</option>
 </select>

编辑:

您可以显式设置值以涵盖循环外部与 || 连接分支值的特殊情况，只要这些值位于返回的 json 数据内即可:

$(document).ready(function(){

       $("[name=ed_job_location]").val(json[0].branches_id + "||" + json[0].branches_name + "||" + json[0].branches_code);

    });