發送ajax請求時PHP不返回任何數據

Question

當我向我的 getMessages.php 文件發送 ajax post 請求時，它不返回任何內容。

我試過手動設置數組值並在控制台中打印它們，這似乎有效。

獲取消息.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來進行測試。

Answer 1

數據在循環中一次又一次地被覆蓋，我猜你想做這樣的事情：

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

Answer 2

哎呀！

我忘記開始會話了！

謝謝帕特里克Q！

Answer 3

這將解決問題，因為您將響應作為對象返回

更新：

You will need to initialize session

數組的數據參數應該在 if 語句中

試試下面的代碼

<?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。 這行代碼可以正常工作

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

Answer 4

您好，您可以嘗試以下方法：

php腳本：

<?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);
})

希望能幫助到你