Question

好吧所以我使用的是PHP，我需要将文本框的内容放入我的数据库表中，当我点击提交时，不会抛出错误，但是这里没有任何内容放在我的代码中：

HTML：

<html>
<title>Chat App</title>
<head>
    <link rel="stylesheet" type="text/css" href="chatCSS.css">
</head>
<body>
    <div class="logOut">
        <form action="loggedOut.php" method="POST">
            <input type="submit" value="Log Out">
        </form>
    </div>
    <div class="incomeMessages">
        <input type="text" id="incomeMessageBox" disabled="disabled">
        <br>
        <br>
        <form method="POST" action="index.php">
            <input type="text" name="message" id="myMessage" placeholder="Message here...">
            <input type="submit" id="myMessageSubmit">
        </form>
    </div>
</body>
</html>

PHP：

<?php
require 'includes/connect_mysql.php';
require 'includes/sessionCheck.php';

if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $message = $_POST['message'];
    $time = time();
    $user = $_SESSION['activeUser'];

    if ($db_connect) {
        $SQL = "INSERT INTO mainconversation (Name, Message, Time) VALUES ('$user', '$message', '$time')";

        $result = mysqli_query($connection, $SQL);
        mysqli_close($connection);
    }

}

?>

连接代码：

<?PHP

$db_host = "localhost"; 

$db_username = "xxxx";  

$db_pass = "xxxxxxxx";  

$db_name = "chatapp"; 


$connection = mysqli_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
$db_connect = mysqli_select_db($connection, "$db_name") or die ("no database");  

?>

Answer 1

您没有检查错误，请尝试更改您的行$结果：

$result = mysqli_query($connection,$SQL) or die(mysqli_error($connection));

如果存在错误，您将收到错误。

Answer 2

得到它，$user变量已经包含' '，所以当我把它们放在SQL中的变量时，它将它作为" "

感谢所有帮助!!!

无法使用PHP INSERT INTO数据库，不会抛出任何错误

2 个答案: