我花了将近两天的时间来寻找我的问题的解决方案，但没有找到任何答案。可能是我最近开始学习 Ajax 和 jquery 的东西，我错过了一些东西。

注意:我的所有选择框都是已选多选框，所有选项都是硬编码的。

我有多个文本和选择框，我想在其中显示/选择来自 mysqli json 输出的数据，这是一个选择框更改事件之一(基本上使用此用户可以复制/编辑保存在数据库中的所有文本的现有详细信息并选择箱)。

我已经成功地使用以下 js 代码对文本框执行此操作，这是我在过去两天学习的以完成我的任务

$(document).ready(function () 
{
   $('#posted_jobs').on('change', function () 
   {
      var selectedValue = $(this).val();    
      $.ajax
      ({
          url: 'Copyvalue.php',
          type: 'POST',
          data: 'filter_job_id=' + selectedValue,
          success: function(response) 
          {
            var obj = $.parseJSON(response);//parse JSON            
            console.log(response);
            //alert(JSON.stringify(obj));       
            $("#Job_Type").val(["Normal"]); // have tried manually adding value to selectbox here but not working
            $('#Locations').val(obj[0].Job_type); // have tried manually adding value to selectbox here but not working
            $('#Job_Type').val(obj[0].Job_type);
            $('#Keywords').val(obj[0].Keywords);
            $('#designation').val(obj[0].Designation);
            $('#Company_name').val(obj[0].Company_Name);
            $('#Contact_Person').val(obj[0].Contact_person_name);
            $('#Contact_person_no').val(obj[0].Contact_No);
            $('#Job_name').val(obj[0].Job_name);
           }
        });         
     });
  });

我已经尝试了 n 个来自 stack-overflow 和其他网络建议的选项。但无法使它适用于选择框。请帮助我完成这项工作，如果可能的话，请告诉我从哪里可以学到的好资源。

尝试过类似的选项

$("#Locations").val(["1, 2, 3, 4"]).prop("selected", false);
$("#Locations").val([1, 2]);

我在 Chrome 控制台中输出的 PHP 脚本 Json

 [{"Job_id":"6","Employer_id":"2","creation_on":"2015-01-03 18:48:58","Keywords":"operation executive, manager operation, operation manager, executive operation","Job_type":"Normal","Designation":"Assistant Manager - Operation","Open_Positions":"4","Job_Description":"<p><strong>Good<\/strong> <em>Position<\/em><\/p>","Min_age":"23","Max_age":"34","Min_exp":"5","Max_exp":"12","Min_salary":"104","Max_salary":"109","Hide_Salary":"","Locations":"3, 4, 8, 45, 46","Industry":"10002, 10004","FA":"1002, 1003","Education":"4, 6, 9","Company_Name":"Aviva Life Insurance Company India Limited","About_Company":"<p><strong>Good<\/strong><em> Company<\/em><\/p>","Contact_person_name":"Balvinder Rayat","Contact_No":"9818906677","Refresh_type":"15","Response_type":"VC","Job_name":"Operation AM","Job_status":"Active"}]

我的选择框

<select id="Locations" name='Locations[]' style='width:100%;' data-placeholder='Type or select   
locations' multiple class='chosen-select-no-results'>
    <option value="1">Bangalore</option>
    <option value="2">Chennai</option>
    <option value="3">Delhi</option>
    <option value="4">Gurgaon</option>
    <option value="5">Hyderabad</option>
    <option value="6">Kolkata</option>
    <option value="7">Mumbai / Navi Mumbai</option>
</select>                      

最佳答案

我在填充 <select> 时也遇到了类似的问题使用 jQuery 选择的插件并将一些选项设置为默认选择。 我在代码中使用的方法是，

• 从 ajax 调用中获取 JSON 响应
• 填充 <select>使用 JSON 的元素
• 遍历 <option>元素
• 使用“selected”属性更新它们(针对特定选项)

我使用的代码片段如下。

var response = "[{\"id\":\"1\",\"location\":\"Chennai\"},{\"id\":\"2\",\"location\":\"Bangalore\"},{\"id\":\"3\",\"location\":\"Hyderabad\"},  {\"id\":\"4\",\"location\":\"Goa\"},{\"id\":\"5\",\"location\":\"Delhi\"},{\"id\":\"6\",\"location\":\"Mumbai\"},{\"id\": \"7\",\"location\":\"Cochin\"},{\"id\":\"8\",\"location\":\"Trivandrum\"}]";

/*populate <select> with the list of locations*/
var locationList = JSON.parse(response);
$.each(locationList, function() {
  $("#location-select").append($("<option/>")
    .val(this.id).text(this.location));
});
$("#location-select").chosen({
  no_results_text: "Oops, no location found!",
  display_selected_options: false
});

/*Make some locations from the list to be selected by default*/
var selectedLoc = JSON.parse("[{\"id\":\"2\",\"location\":\"Bangalore\"},{\"id\":\"6\",\"location\":\"Mumbai\"},{\"id\": \"7\",\"location\":\"Cochin\"}]");

/*
 *Create an array of locations, it can be array of id also,then
 * the if condition should be
 * if ($.inArray($(this).val(), users) != -1){...}
 * This is because id is populated as
 * <select>
 *  <option value="id from json"> location from json </option>
 * </select>
 */
var locations = [];
$.each(selectedLoc, function() {
  locations.push(this.location);
});

$("#location-select option").each(function() {
  if ($.inArray($(this).text(), locations) != -1) {
    $(this).attr('selected', 'selected');
    $("#location-select").trigger('chosen:updated');
    return;
  }
});

<link href="https://cdnjs.cloudflare.com/ajax/libs/chosen/1.6.2/chosen.min.css" rel="stylesheet" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/chosen/1.6.2/chosen.jquery.min.js"></script>
<h3>
jQuery Chosen multiselect with selected values by default
</h3>
<select id="location-select" data-placeholder="Select locations..." style="width:95%;" multiple class="chosen-select"></select>

关于php - 将选择的多选选项值设置为使用 json 选择，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27875492/