好吧，我正在使用 PHP，我需要将文本框的内容放入数据库的表中，当我单击“提交”时，不会抛出错误，但数据库中不会放入任何内容，这是我的代码:
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");
?>
最佳答案
您没有检查错误，请尝试更改您的 $result 行，如下所示:
$result = mysqli_query($connection,$SQL) or die(mysqli_error($connection));
如果有错误，您将会收到错误。
关于php - 无法使用 PHP INSERT INTO 数据库，没有抛出错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24871929/