发送ajax请求时PHP不返回任何数据

Question

当我向我的 getMessages.php 文件发送 ajax post 请求时，它不返回任何内容。

我试过手动设置数组值并在控制台中打印它们，这似乎有效。

获取消息.php

<?php

require_once "mysqli.php";

$data = array();

if (isset($_POST['getChat']) && !empty($_POST['getChat'])) {
    $username = $_SESSION["username"];

    $result = mysqli_query($conn, "SELECT msg_startuser, msg, time 
        FROM messages 
        WHERE msg_startuser = '{$username}' and msg_enduser = 'mariokiller470' 
        UNION 
        SELECT msg_startuser, msg, time 
        From messages 
        WHERE msg_startuser = 'mariokiller470' and msg_enduser = '{$username}' 
        order by time;
    ");

    while ($row = mysqli_fetch_array($result)) {
        $data['startuser'] = $row['msg_startuser'];
        $data['msg'] = $row['msg'];
    }
}

echo json_encode($data);
exit;
?>

js ajax

function getChat() {

    $.ajax({
        url: 'getMessages.php',
        type: 'POST',
        data: {getChat: 'yes'},
        dataType: 'JSON',
        success: function(data) {
            // testing
            console.log(data.startuser, data.msg);
        }

    })

}

我希望它在控制台中打印出来进行测试。

Answer 1

数据在循环中一次又一次地被覆盖，我猜你想做这样的事情：

$x = 0;
while ($row = mysqli_fetch_array ($result)) {
        $data[$x]['startuser'] = $row['msg_startuser'];
        $data[$x]['msg'] = $ row['msg'];
        $x++;
}

Answer 2

哎呀！

我忘记开始会话了！

谢谢帕特里克Q！

Answer 3

这将解决问题，因为您将响应作为对象返回

更新：

You will need to initialize session

数组的数据参数应该在 if 语句中

试试下面的代码

<?php

require_once "mysqli.php";
session_start();

if (isset($_POST['getChat']) && !empty($_POST['getChat'])) {
    $username = $_SESSION["username"];

$data = array();


    $result = mysqli_query($conn, "SELECT msg_startuser, msg, time 
        FROM messages 
        WHERE msg_startuser = '{$username}' and msg_enduser = 'mariokiller470' 
        UNION 
        SELECT msg_startuser, msg, time 
        From messages 
        WHERE msg_startuser = 'mariokiller470' and msg_enduser = '{$username}' 
        order by time;
    ");

    while ($row = mysqli_fetch_array($result)) {
         $startuser = $row['msg_startuser'];
        $msg = $row['msg'];

$data = array("startuser" =>$startuser, "msg" =>$msg);
    }


echo json_encode($data);
exit;

}
?>

所以在ajax控制台中。 这行代码可以正常工作

console.log(data.startuser, data.msg);

Answer 4

您好，您可以尝试以下方法：

php脚本：

<?php
require_once "mysqli.php";
session_start();// start the session
$data = array();
if (isset($_SESSION["username"])) {
  if (isset($_POST["endUser"]) &&  isset($_POST["action"])) {
  $case = $_POST["action"];
  $endUser = $_POST["endUser"];
  $username = $_SESSION["username"];
  switch (case) {
    case 'getChat':
    $result = mysqli_query($conn, "SELECT msg_startuser, msg, time
        FROM messages
        WHERE msg_startuser = '{$username}' and msg_enduser = '{$endUser}'
        UNION
        SELECT msg_startuser, msg, time
        From messages
        WHERE msg_startuser = '{$endUser}' and msg_enduser = '{$username}'
        order by time;
    ");

    while ($row = mysqli_fetch_assoc($resultado)) {
        if (isset($row['msg_startuser']) && isset($row['msg'])) {
          $temp = array(
            "user"=>$row['msg_startuser'],
            "msg"=>$row['msg']
          );
        }
        $data[] = $temp;
    }
    echo json_encode($data);
      break;
  }
  }
}else {
  echo "error-403";
}
 ?>

javascript：

 function getChat() {

    return $.ajax({
         url: 'getMessages.php',
         type: 'POST',
         data: {action: 'getChat',endUser:'mariokiller470'},
         dataType: 'JSON'
     })

 }

getChat()
.done(function(response){
  console.log(response);
})

希望能帮助到你