我如何使下拉值使用jsonencode數據使用Javascript選擇

Question

在我的html格式中，我有9個下拉值，如果用戶操作是編輯，那麼它會從數據庫中獲取值並返回jsonencode format.as下面，

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"}] 

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> 

我嘗試了幾行代碼使用PHP其實際工作，但我想使用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> 

所以在這裏老實話，我要提取JSON值並下拉值在頁面加載有「選擇」。

編輯：

JSON數據

[{"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> 

從上面的JSON

在這裏，我只得到分支ID和DESIG ID，但她我對||有價值在選擇選項中的符號，所以我需要找到特定的ID並在下拉菜單中顯示它。

Answer 1

通過json對象數組迭代並通過它的名字中選擇填充每個元素值：

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>

編輯：

可以明確設置的值，以覆蓋只要值在json數據返回範圍內，分支值與||級聯的特殊情況編輯：

$(document).ready(function(){ 

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

    });