如何通過jsonencode數據使用Javascript選擇下拉值

Question

在我的html表單中，我有9個下拉值，如果用戶操作為Edit，則它將從數據庫中獲取值並以jsonencode格式返回。如下所示，

JSON數據

 [{"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"}] 

的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> 

我嘗試了幾行使用php的代碼，它實際上有效，但是我正在嘗試使用javascript。

的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> 

因此，在這里，我必須提取json值，並使下拉值在頁面加載時具有“已選擇”狀態。

編輯：

JSON數據

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

的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>

這里從json上面，我只獲得分支ID和設計ID，但是這里我具有||的價值。 選擇選項中的符號，因此我需要找到特定的ID並在下拉菜單中顯示它。

Answer 1

遍歷json對象數組，並通過其名稱選擇每個元素值來填充它：

 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> 

編輯：

您可以顯式設置值，以使用||覆蓋分支值串聯的特殊情況。 在循環外部，只要值在返回的json數據之內：

$(document).ready(function(){

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

    });