如何在AJAX方法中将数据链接到选择元素? [英] How can I link data to a select element in AJAX method?

问题描述

我有一个问题，我无法将数据放入select元素中，也无法使用ID做出附加到我的ajax内部内容的选项.我得到了数据，它显示在输入元素中，但是当我将其切换到选择元素时，它不起作用.

这是我表格的图像

JQuery/Ajax代码

 函数ToolsChange(element){让tools_id = $(element).val();如果(tools_id){$ .ajax({类型:"post"，网址:"form_JSON_approach.php"，数据: {"tools_id":tools_id}，成功:功能(响应){var dataSplit =响应；console.log(response);var shouldSplit = dataSplit.split("@");var shouldNotSplit = dataSplit.split();console.log(shouldSplit);console.log(shouldSplit [0]);console.log(shouldSplit [1]);console.log(shouldSplit [2]);$(#sel_control_num").val(shouldSplit [0]);var specs = [];对于(i = 1; i< shouldSplit.length; i + = 3){specs.push(shouldSplit [i])}$(#sel_tools_spec").val(specs.join('，'));$(#sel_tools_id").val(shouldSplit [2]);}});}} 

HTML代码(我必须评论select元素，因为它没有显示数据)

 < div class ="form-group"><标签>工具规范:</label>< input id =" sel_tools_spec"class ="form-control"name ="tools_specification"data-live-search ="true"readonly><！-< select id =" sel_tools_spec"class ＝"form-control selectpicker".data-live-search ="true"</select>-></div> 

PHP代码

 <?phpinclude("../include/connect.php");if(isset($ _ POST ['tools_id'])){$ ID = $ _POST ['tools_id'];$ query =" SELECT tools_masterlist.control_no，tools_masterlist.tools_id，tools_masterlist.tools_name，tools_spec.model_num，tools_spec.model_num_val，tools_spec.status从tools_masterlist左加入tools_spec ON tools_masterlist.tools_id = tools_spec.tools_id左联接tools_registration ON tools_masterlist.control_no = tools_registration.reg_input状态= 1 AND tools_name ='$ ID';$ con-> next_result();//$ result = mysqli_query($ con，" CALL GetAjaxForToolsRegistration('$ ID')");;$ result = mysqli_query($ con，$ query);if(mysqli_num_rows($ result)> 0){while($ row = mysqli_fetch_assoc($ result)){//echo $ row ['control_no']."@".$ row ['model_num']."@".$ row ['tools_id'].echo $ row ['control_no']."@".'< option value ='''.$ row ['tools_id'].'>'.$ row ['model_num'].'</option>'."@".$ row ['tools_id'].}}别的{}}?> 

解决方案

您要一次附加所有数据，而不必在for循环中直接在 selectpicker 内部附加选项并刷新.>

演示代码 :

  $(#sel_tools_spec").selectpicker()//在加载时初始化ToolsChange()//仅用于演示..函数ToolsChange(element){/*让tools_id = $(element).val();如果(tools_id){$ .ajax({类型:"post"，网址:"form_JSON_approach.php"，数据: {"tools_id":tools_id}，成功:功能(响应){*///其他代码...$(#sel_tools_spec").html('');//假设数据看起来像这样...var shouldSplit = ["1"，< option> A</option>"，"1001"，"2"，< option> B</option>"，"1001"]对于(i = 1; i< shouldSplit.length; i + = 3){//在选择框内附加选项$(#sel_tools_spec").append(shouldSplit [i]);}$(#sel_tools_spec").selectpicker('refresh');//刷新/*}}); */}  

 <链接rel ="stylesheet" type ="text/css" href ="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">< link rel ="stylesheet" href ="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-select/1.12.2/css/bootstrap-select.min.css">< script src ="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.js"></script>< script src ="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"</script>< script src ="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-select/1.12.2/js/bootstrap-select.min.js"</script>< div class ="form-group"><标签>工具规范:</label>< select id ="sel_tools_spec" class ="form-control selectpicker" data-live-search ="true"></select></div>  

I have a problem wherein I cannot put the data inside select element and make an option using the ID to append on what is inside my ajax. I got the data and it is showing in an input element but when I switched it into select element it doesn't work.

Here is the image of my form

JQuery / Ajax code

    function ToolsChange(element) {
    let tools_id = $(element).val();

    if (tools_id) {

        $.ajax({
            type: "post",
            url: "form_JSON_approach.php",
            data: {
                "tools_id": tools_id
            },
            success: function(response) {
                var dataSplit = response;
                console.log(response);
                var shouldSplit = dataSplit.split("@");
                var shouldNotSplit = dataSplit.split();
                console.log(shouldSplit);
                console.log(shouldSplit[0]);
                console.log(shouldSplit[1]);
                console.log(shouldSplit[2]);
                $("#sel_control_num").val(shouldSplit[0]);

                var specs = [];
                for (i = 1; i < shouldSplit.length; i += 3) {
                    specs.push(shouldSplit[i])
                }

                $("#sel_tools_spec").val(specs.join(', '));

                $("#sel_tools_id").val(shouldSplit[2]);
            }
        });
    }
  }

HTML code(I had to comment select element because it is not showing the data)

<div class="form-group">
     <label> Tools Specification: </label>
     <input id="sel_tools_spec" class="form-control" name="tools_specification"
      data-live-search="true" readonly>
<!-- <select id="sel_tools_spec" class="form-control selectpicker" data-live-search="true">
      </select> -->
</div>

PHP code

    <?php 

    include("../include/connect.php");

 
    if(isset($_POST['tools_id'])){
        
        $ID = $_POST['tools_id'];

        $query = "SELECT tools_masterlist.control_no, tools_masterlist.tools_id, 
        tools_masterlist.tools_name, 
        tools_spec.model_num,tools_spec.model_num_val, tools_spec.status
        FROM tools_masterlist LEFT JOIN tools_spec ON tools_masterlist.tools_id = tools_spec.tools_id
        LEFT JOIN tools_registration ON tools_masterlist.control_no = tools_registration.reg_input 
        WHERE status = 1 AND tools_name = '$ID'";

        $con->next_result();
        // $result=mysqli_query($con, "CALL GetAjaxForToolsRegistration('$ID')");
        $result=mysqli_query($con, $query);
        if(mysqli_num_rows($result)>0)
        {
            while($row = mysqli_fetch_assoc($result))
            {
                // echo $row['control_no'] . "@" . $row['model_num'] . "@" . $row['tools_id'] ."@";
                echo $row['control_no'] . "@" . '<option value="'.$row['tools_id'].'">'. 
                $row['model_num'] .'</option>' . "@" . $row['tools_id'] ."@";
            }
        }
        else
        {
          
        }

    }
  ?>

解决方案

You are appending all datas at onces instead inside for-loop you can directly append options inside your selectpicker and refresh it.

Demo Code :

$("#sel_tools_spec").selectpicker() //intialize on load
ToolsChange() //just for demo..
function ToolsChange(element) {
  /*let tools_id = $(element).val();

  if (tools_id) {

    $.ajax({
      type: "post",
      url: "form_JSON_approach.php",
      data: {
        "tools_id": tools_id
      },
      success: function(response) {*/
  //other codes....
  $("#sel_tools_spec").html('');
  //suppose data look like this...
  var shouldSplit = ["1", "<option>A</option>", "1001", "2", "<option>B</option>", "1001"]
  for (i = 1; i < shouldSplit.length; i += 3) {
    //append options inside select-box
    $("#sel_tools_spec").append(shouldSplit[i]);
  }
  $("#sel_tools_spec").selectpicker('refresh'); //refresh it
  /* }
    });*/
}

<link rel="stylesheet " type="text/css" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-select/1.12.2/css/bootstrap-select.min.css">
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-select/1.12.2/js/bootstrap-select.min.js"></script>
<div class="form-group">
  <label> Tools Specification: </label>
  <select id="sel_tools_spec" class="form-control selectpicker" data-live-search="true">
  </select>
</div>

这篇关于如何在AJAX方法中将数据链接到选择元素?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！