為何我連上mysql了卻接收不了表單

mysql匯入問題

不明 2022-02-19 18:31:28 ‧ 1436 瀏覽

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<?php
    $conn=mysqli_connect("localhost","root","root","test");
    $sql = "insert into play valuse('$_POST[Account]','$_POST[Password]')";
    $result = mysqli_query($conn,$sql);
    if($result){
        echo"成功";
    }else{
        echo"失敗";
    }
?>


//[form.php]

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>Document</title>
</head>
<body>  
    <form method="POST" action="crud.php">
        <h3>New Account</h3>
        Account: <input type="text" name="Account"><br>
        Password: <input type="text" name="Password"><br>
        <input type="submit" value="insert">
    </form>
</body>
</html>
在測試時,輸入了account 和password後出現

**"Fatal error: Uncaught mysqli_sql_exception: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'valuse('dhgfh','fgh')' at line 1 in F:\xampp\htdocs\test\crud.php:4 Stack trace: #0 F:\xampp\htdocs\test\crud.php(4): mysqli_query(Object(mysqli), 'insert into pla...') #1 {main} thrown in F:\xampp\htdocs\test\crud.php on line 4"**

新手,一星期編程經驗,求賜教

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不明檢舉

我把這php放在了xampp底下,懶得搬別見怪

haward79 iT邦好手 3 級 ‧ 2022-02-19 19:03:25 檢舉

建議你直接貼程式碼
放圖不好改

不明檢舉

好的,謝謝提議

窮嘶發發發 iT邦高手 1 級 ‧ 2022-02-21 09:31:47 檢舉

INSERT INTO table_name (column1, column2, column3...)
VALUES (value1, value2, value3...);
OR
INSERT INTO table_name
VALUES (value1, value2, value3...);

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