我已經走過這段代碼大量的時間不能發現一個單一的錯誤沒有任何人看到這個腳本的任何錯誤。 它給我以下錯誤: 本地主機當前無法處理此請求。 HTTP錯誤500PHP 500內部錯誤沒有得到現場
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
?>
<?php
session_start();
ob_start();
header("Content-type: application/json");
date_default_timezone_set('UTC');
$db = mysqli_connect('localhost', 'root', 'root', 'chat');
if(mysqli_connect_errno()) {
echo '<p>Error:Could not connect to database.<br>
Please try again later.</p>';
exit;
}
try {
$currentTime = time();
$session_id = session_id();
$lastPoll = isset($_SESSION['last_poll']) ?
$_SESSION['last_poll'] : $currentTime;
$action = isset($_SERVER['REQUEST_METHOD']) &&
($_SERVER['REQUEST_METHOD'] == 'POST') ?
'send' : 'poll';
switch($action){
case 'poll':
$query = "SELECT * FROM chatlog WHERE
date_created >= ?";
$stmt = $db->prepare($query);
$stmt->bind_param('s', $lastPoll);
$stmt->execute();
$stmt->bind_result($id, $message, $session_id, $date_created);
$result = $stmt->get_result();
$newChats = [];
while($chat = $result->fetch_assoc()){
if($session_id == $chat['sent_by']){
$chat['sent_by'] = 'self';
}else{ //good
$chat['sent_by'] = 'other';
}
$newChats[] = $chat;
}
$_SESSION['last_poll'] = $currentTime;
print json_encode([
'success' => true,
'messages' => $newChats
]);
exit;
case 'send':
$message = isset($_POST['message']) ? $_POST['message'] : '';
$message = strip_tags($message);
$query = "INSERT INTO chatlog(message, sent_by, date_created)
VALUES(?, ?, ?)";
$stmt = $db->prepare($query);
$stmt->bind_param('ssi', $message, $session_id, $currentTime);
$stmt->execute();
print json_encode(['success' => true]);
exit;
}
}catch(\Exception $e) {
print json_encode([
'success' => false,
'error' => $e->getMessage();
]);
}
我有javascript文件和JavaScript文件通過處理Ajax請求,我使用jQuery來處理所有的AJAX的東西的HTML文件。我不知道是否可以將錯誤傳遞到PHP文件中?
擺脫你的第一個php結束標記和第二個php開放標記,因爲它們之間會有一個空格字符輸出,然後你試圖在任何輸出之前設置需要完成的標題: ) – flauntster