Question

我一直在尝试在将查询发送到数据库之前准备一个查询，但它没有从数据库中取出任何内容。它告诉我user / pass组合不正确。（当它是正确的时），这必须意味着它有来自数据库的0行。任何人都可以告诉我如何解决这个问题吗？

/* Create a prepared statement */
        $stmt = mysqli_stmt_init($con);

        mysqli_stmt_prepare($stmt, "SELECT * FROM users WHERE username = ? AND userpass = ?");

        /* Bind parameters
        s - string, b - blob, i - int, etc */
        mysqli_stmt_bind_param($stmt, "ss", $username, $userpass);

        /* Execute it */
        mysqli_stmt_execute($stmt);

        /* Bind results */
        mysqli_stmt_bind_result($stmt, $username, $userpass);

        /* Fetch the value */
        mysqli_stmt_fetch($stmt);

        /* close statement */
        mysqli_stmt_close($stmt); 


        /* $result = mysqli_query($con,"SELECT * FROM users 
            WHERE username = '$username' AND userpass = '$userpass'"); */


        if(!$stmt)  
        {  
            //something went wrong, display the error  
            echo '<ul class="ulstylecenter">
            <li>Something went wrong while signing in. Please try again later.</li>
            <li>If you are not redirected in 5 seconds please <a href="/home.php">click here</a>.</li>
            </ul>';  
            header('Refresh: 5;url=/home.php');
            //echo mysqli_error(); //debugging purposes, uncomment when needed 
        } 
        else 
        { 
            //the query was successfully executed, there are 2 possibilities 
            //1. the query returned data, the user can be signed in 
            //2. the query returned an empty result set, the credentials were wrong 
            if(mysqli_num_rows($stmt) == 0) 
            { 
                echo '<ul class="ulstylecenter">
                <li>You have supplied a wrong user/password combination. Please try again.</li></ul>';
                echo '<form method="post" action="">  
                        <table>
                        <tr>
                        <th><label for="username" class="signinlabel">Username:</label></th>
                        <td><input type="text" name="username" value="';
                        if(isset($username)){ echo $username; }
                        echo '" class="signininput"></td>
                        </tr>

                        <tr>  
                        <th><label for="userpass" class="signinlabel">Password:</label></th>
                        <td><input type="password" name="userpass" class="signininput"></td>
                        </tr> 
                        </table>
                        <ul class="forgotsignin"><li><a href="#">Forgot your username or password?</a></li></ul>
                        <input type="submit" value="Sign In" class="signinbutton" id="signinbuttonid">  
                </form>';
            } 
            else 
            { 
            $_SESSION['signed_in'] = true; 

                //we also put the user_id and user_name values in the $_SESSION, so we can use it at various pages 
                while($row = mysqli_fetch_assoc($stmt)) 
                { 
                    $_SESSION['user_id']   = $row['user_id']; 
                    $_SESSION['username']  = $row['username']; 
                    $_SESSION['useremail'] = $row['useremail']; 
                    $_SESSION['userdate'] = $row['userdate'];
                } 
                echo '<ul class="ulstylecenter">
                    <li>Succesfully logged in.</li>
                    <li>If you are not redirected in 5 seconds please <a href="/home.php">click here</a>.
                    </li></ul>'; 
                header('Refresh: 5;url=/home.php');
            }

Answer 1

代码中存在很多缺陷。只有其中一些：

您需要添加mysqli_stmt_bind_result参数变量，这些变量将代表表users的所有列。例如，如果您有表字段：
```
id    username    password    email    date
```
您的mysqli_stmt_bind_result电话应如下所示：
```
mysqli_stmt_bind_result($stmt, $id, $username, $userpass, $email, $date);
```
因为查询包含SELECT *。
请勿在{{1}}之后检查if (!$stmt)。
如果您已经有mysqli_stmt_close($stmt);和$username值，则无法再从数据库中检索它们......

强烈建议仔细阅读整个mysqli documendation。

我认为你的代码应该是这样的（如果你更喜欢程序化的mysqli风格）：

$userpass

Answer 2

使用var_dump（）检查变量，然后在这行之后尝试设置变量$ username和$ userpass ：

mysqli_stmt_bind_result($stmt, $username, $userpass);

PHP准备查询不起作用

2 个答案: