在核心php中获取索引错误..我使用ajax和jquery

Question

< script >
  $(document).ready(function() {

    $(document).on('keydown', '.username', function() {

      var id = this.id;
      var splitid = id.split('_');
      var index = splitid[1];

      // Initialize jQuery UI autocomplete
      $('#' + id).autocomplete({
        source: function(request, response) {
          $.ajax({
            url: "getDetails.php",
            type: 'post',
            dataType: "json",
            data: {
              search: request.term,
              request: 1
            },
            success: function(data) {
              response(data);
            }
          });
        },
        select: function(event, ui) {
          $(this).val(ui.item.label); // display the selected text
          var userid = ui.item.value; // selected value

          // AJAX
          $.ajax({
            url: 'getDetails.php',
            type: 'post',
            data: {
              userid: userid,
              request: 2
            },
            dataType: 'json',
            success: function(response) {

              var len = response.length;

              if (len > 0) {
                var id = response[0]['id'];
                var name = response[0]['name'];
                var email = response[0]['email'];
                var age = response[0]['age'];
                var salary = response[0]['salary'];

                // Set value to textboxes
                document.getElementById('name_' + index).value = name;
                document.getElementById('age_' + index).value = age;
                document.getElementById('email_' + index).value = email;
                document.getElementById('salary_' + index).value = salary;

              }

            }
          });

          return false;
        }
      });
    });

    // Add more
    $('#addmore').click(function() {

      // Get last id 
      var lastname_id = $('.tr_input input[type=text]:nth-child(1)').last().attr('id');
      var split_id = lastname_id.split('_');

      // New index
      var index = Number(split_id[1]) + 1;

      // Create row with input elements
      var html = "<tr class='tr_input'><td><input type='text' class='username' id='username_" + index + "' placeholder='Enter username'></td><td><input type='text' class='name' id='name_" + index + "' ></td><td><input type='text' class='age' id='age_" + index + "' ></td><td><input type='text' class='email' id='email_" + index + "' ></td><td><input type='text' class='salary' id='salary_" + index + "' ></td></tr>";

      // Append data
      $('tbody').append(html);

    });
  }); <
/script>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<div class="container">

  <table border='1' style='border-collapse: collapse;'>
    <thead>
      <tr>
        <th>Username</th>
        <th>Name</th>
        <th>Age</th>
        <th>Email</th>
        <th>Salary</th>
      </tr>
    </thead>
    <tbody>
      <tr class='tr_input'>
        <td><input type='text' class='username' id='username_1' placeholder='Enter username'></td>
        <td><input type='text' class='name' id='name_1'></td>
        <td><input type='text' class='age' id='age_1'></td>
        <td><input type='text' class='email' id='email_1'></td>
        <td><input type='text' class='salary' id='salary_1'></td>
      </tr>
    </tbody>
  </table>
  <br>
  <input type='button' value='Add more' id='addmore'>
</div>
<?php
include "config.php";

$request = $_POST['request']; // request

// Get username list
if($request == 1){
 $search = $_POST['search'];

 $query = "SELECT * FROM users WHERE username like'%".$search."%'";
 $result = mysqli_query($con,$query);
 
 while($row = mysqli_fetch_array($result) ){
  $response[] = array("value"=>$row['id'],"label"=>$row['username']);
 }

 // encoding array to json format
 echo json_encode($response);
 exit;
}

// Get details
if($request == 2){
 $userid = $_POST['userid'];
 $sql = "SELECT * FROM users WHERE id=".$userid;

 $result = mysqli_query($con,$sql); 

 $users_arr = array();

 while( $row = mysqli_fetch_array($result) ){
  $userid = $row['id'];
  $fullname = $row['fname']." ".$row['lname'];
  $email = $row['email'];
  $age = $row['age'];
  $salary = $row['salary'];

  $users_arr[] = array("id" => $userid, "name" => $fullname,"email" => $email, "age" =>$age, "salary" =>$salary);
 }

 // encoding array to json format
 echo json_encode($users_arr);
 exit;
}

my database CREATE TABLE `users` (
  `id` int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
  `username` varchar(80) NOT NULL,
  `fname` varchar(60) NOT NULL,
  `lname` varchar(60) NOT NULL,
  `email` varchar(80) NOT NULL,
  `age` int(2) NOT NULL,
  `salary` varchar(10) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

得到索引错误...它是核心php从数据库获取值...它无法从数据库加载值

Answer 1

你可以这样对

if (!empty($_POST)) 
{ 
    $search = $_POST['search']; 
    $query = "SELECT * FROM users WHERE username like'%".$search."%'"; $result = mysqli_query($con,$query); 

    while($row = mysqli_fetch_array($result))
    {
        $response[] = array("value"=>$row['id'],"label"=>$row['username']);
    }echo json_encode($response);
    exit;
}

Answer 2

您正在使用mysqli_fetch_array，但之后您尝试访问列名称（例如$row['fname']）的列。为此，您必须使用mysqli_fetch_assoc。

使用mysqli_fetch_array，您只能使用列索引（$row[2]，$row[3] ...例如）