Question

我有一个网站，我想将新学生添加到拼写网站，但由于某种原因，添加学生的代码不会添加学生。我确定我的查询很好，因为我从phpMyAdmin生成的示例中复制了它。

connection.php

<?php

define("DB_SERVER", "localhost");
define("DB_USER", "root");
define("DB_PASS", "root");
define("DB_NAME", "spelling");

$connection = new mysqli(DB_SERVER,DB_USER,DB_PASS,DB_NAME);

if (mysqli_connect_errno()) {
    exit("Database connection failed: " . mysqli_connect_error());};
?>

addstudent.php

<?php session_start();
require_once("connection.php");
?>
<!DOCTYPE html>
<html>
<head>
    <title>
    Add a student
    </title>
</head>
<body>
<form action="addstudent.php" method="post">
        Full Name:
        <input type="text" name="fullname" autocomplete="off">
        <br />
        Username: 
        <input type="text" name="username" autocomplete="off">
        <br />
        Password:  
        <input type="password" name="password" autocomplete="off">
        <br />
        <br />
        <input type="submit" value="Add Student" name="submit">
        </form>
<?php

if(isset($_POST["submit"]) ){
    $_POST = array_map("strip_tags",$_POST);
    $_POST = array_map("trim",$_POST);

    $fullname = $_POST["fullname"];
    $username = $_POST["username"];
    $password = $_POST["password"];

    $query = "INSERT INTO `spelling`.`students` (`SID`, `fullname`, `CID`, `username`, `password`) VALUES (NULL, '".$fullname."', '1', '".$username."', '".$password."')";
    $result = mysqli_query($query, $connection);
    }
?>
</body>
</html>

如果代码在语法上是正确的，我的问题可能在哪里？

Answer 1

$result = mysqli_query($query, $connection);

您需要在此行上切换查询和连接。像这样：

$result = mysqli_query($connection, $query);

Answer 2

问题出在$result = mysqli_query($query, $connection);

您切换 $ query和$ connection，see the reference

mysqli_query ( mysqli $link , string $query ... )

Answer 3

我注意到您在一个看似是（主）键（SID）的列中输入NULL。也许这就是问题所在？

PHP和MySQL代码不插入新行

3 个答案: