Question

我有一个问题，什么时候不能运行这个程序，它显示

解析错误：语法错误，意外的'用户名'（T_STRING）第9行的C：\ xampp \ htdocs \ fypp \ index.php

<?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，正确粘贴代码......最后您遗漏了</table>和</form> ...

其次，你错过了一行中的简单撇号和分号

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

此外，您应该使用mysqli_query代替mysqlui_query ...错字！

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程序，而且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，您错过了同一行的半冒号（;）终止。

PHP网站调试

2 个答案: