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Question

我有一个问题，什么时候不能运行这个程序，它显示

Parse error: syntax error, unexpected 'username' (T_STRING) in C:\xampp\htdocs\fypp\index.php on line 9

<?php
    session_start();
    if (isset($_POST['bttLogin'])){
        require 'connect.php';
        $username = $_POST['username'];
        $password = $_POST['password'];
        $result = mysquli_query($con, 'select * from account where username="'.$username.'" and password="'.$password.'")
        if(mysqli_num_rows($result)==1) {
            $_SESSION['username'] = $username;
            header("Location: welcome.php");
        }
        else
            echo "account is invalid";
    }
    ?>
    <form method="post">
    <table cellpadding="2" cellspacing="2" border="1">
        <tr>
            <td>Username</td>
            <td><input type="text" name="username"></td>
        </tr>
        <tr>
            <td>Password</td>
            <td><input type="password"" name="password"></td>
        </tr>
        <tr>
            <td>&nbsp;</td>
            <td><input type="submit" name="bttLogin" value="Login"></td>
        </tr>

感谢您的回答:)

Answer 1

首先，正确粘贴您的代码...您最后缺少 </table> 和 </form>...

其次，您在

行中遗漏了一个简单的撇号和一个 semi-colon

$result = mysquli_query($con, 'select * from account where username="'.$username.'" and password="'.$password.'")

此外，您应该使用 mysqli_query 而不是 mysqlui_query... 打字错误！

您的代码应如下所示：

<?php
session_start();
if (isset($_POST['bttLogin'])){
    require 'connect.php';
    $username = $_POST['username'];
    $password = $_POST['password'];
    $result = mysqli_query($con, 'select * from account where username="'.$username.'" and password="'.$password.'"');
    if(mysqli_num_rows($result)==1) {
        $_SESSION['username'] = $username;
        header("Location: welcome.php");
    }
    else
        echo "account is invalid";
}
?>

<form method="post">
<table cellpadding="2" cellspacing="2" border="1">
    <tr>
        <td>Username</td>
        <td><input type="text" name="username"></td>
    </tr>
    <tr>
        <td>Password</td>
        <td><input type="password"" name="password"></td>
    </tr>
    <tr>
        <td>&nbsp;</td>
        <td><input type="submit" name="bttLogin" value="Login"></td>
    </tr>
</table>
</form>

Answer 2

将 mysquli_query 更改为 mysqli_query，因为它不是有效的扩展名，我认为它的拼写错误。

修改后的查询：

$result = mysqli_query($con, "SELECT * FROM account 
                          WHERE username='" . $username . "' 
                          AND password ='". $password ."'");

另请注意，您错过了同一行的终止符 分号 (;)。

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