Question

我会尽力解释这个好事。当我通过ajax btw提交数据时，我在action.php中的变量始终为null我试图使用模态编辑我的数据。我只是不知道我哪里出错了我希望你们明白我的观点。这是我的代码......

// html表单

<div class="modal fade" id="exampleModalLong" tabindex="-1" role="dialog" aria-labelledby="exampleModalLongTitle" aria-hidden="true">
          <div class="modal-dialog" role="document">
            <div class="modal-content">
              <div class="modal-header">
                <h2 class="modal-title" id="exampleModalLongTitle">Edit Fields</h2>
              </div>
              <div class="modal-body">
                <form id="form_data">
                    <form id="edit_data">
                        <input type="text" name="id" id="id2">
                        <input type="text" name="username" id="username2" class="form-control"><br>
                        <input type="text" name="subject" id="subject2" class="form-control"><br>
                        <input type="text" name="plan" id="plan2" class="form-control"><br>
                        <input type="submit" name="go_edit" value="Edit" id="edit2" class="btn btn-warning">
                    </form>
                </form>
              </div>

            </div>
          </div>
        </div>

//这是ajax脚本

$('#edit2').on('click', function(event){
            event.preventDefault();
            var edit = $("#edit_data").serialize() + "&go_edit=1";
            $.ajax({
                url:"action.php",
                method:"POST",
                data:edit,
                success:function(data2){
                alert(data2);
                $('#exampleModalLong').modal('hide');
                $('#list').load(window.loccation = ' #list');
                }
                });

            });
        });

// php代码，用于编辑位于action.php

中的表单中的数据

if(isset($_POST['go_edit'])){
    $id = $_POST['id'];
    $username = $_POST['username'];
    $subject = $_POST['subject'];
    $plan = $_POST['plan'];
    var_dump($where = array("id" => $id));
    var_dump($myArray = array("username" => $username,
                    "subject" => $subject,
                    "plan" => $plan));
    if($data->EditData("new_users",$myArray,$where)){

        echo "Data Edited Successfully";
    }
}

//这是我在edit.php

中编辑的sql

public function EditData($table,$fields,$where){
            $sql = "";
            $condition = "";

            foreach ($where as $key => $value) {
                $condition .= $key ."='". $value. "' AND ";
            }
            $condition = substr($condition, 0, -5);

            foreach ($fields as $key => $value) {
                $sql .= $key . "='".$value."', ";
            }
            $sql = substr($sql, 0, -2);
            $sql = "UPDATE ".$table." SET ".$sql." WHERE ".$condition;

            $query = mysqli_query($this->con,$sql);

            if($query){
                return true;
            }

        }

Answer 1

对jQuery代码进行一些修改：获取所有值：

var id1 = $("#id2").val();
var un1 = $("#username2").val();
var s1 = $("#subject2").val();
var p1= $("#plan2").val();
var submit=$('#edit2').val();

使用数据异步调用PHP代码：

$.post("action.php", {
 id: id1,
 username: un1,
 subject: s1,
 plan: p1,
 go_edit:submit
},function(data) {
$("#returnmessage").append(data); // Append returned message to message
if (data == "TRUE") {
   $("#form")[1].reset(); // To reset form fields on success.
}

为什么我的变量在使用ajax发送数据时始终为null

1 个答案: