mysqli_stmt::bind_result()：绑定变量的数量与 PHP 中准备好的语句中的字段数量不匹配 [重复]

Question

我正在尝试进行用户名搜索 [我似乎已经完成并且工作正常]，但是当您搜索用户名时，将显示有关该帐户的信息。比如我搜索过virtualAnon，他的名字和first_name等信息会出现在他的用户名后面。

我尝试通过将$query = "SELECT username FROM users WHERE username like ? LIMIT 1"; 替换为$query = "SELECT * FROM users WHERE username like ? LIMIT 1"; 来修复它，但在我尝试之后，错误

mysqli_stmt::bind_result(): 绑定变量数不匹配 PHP中准备好的语句中的字段数

出现了。

这是获取用户名和数据库的 PHP 文件：

<?php
    if($_GET['keyword'] && !empty($_GET['keyword']))
    {
        $conn = mysqli_connect('localhost','root','','loginsecure'); //Connection to my database
        $keyword = $_GET['keyword'];
        $search = $_GET['keyword'];
        $keyword="%$keyword%";
        $query = "SELECT * FROM users WHERE username like ? LIMIT 1";
        # When I tried to SELECT *, It gives me the error of: Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in ...\fetch.php on line 22
        $statement = $conn->prepare($query);
        $statement->bind_param('s',$keyword);
        $statement->execute();
        $statement->store_result();
        if($statement->num_rows() == 0) // so if we have 0 records acc. to keyword display no records found
        {
            echo '<div id="item">Sorry, but there is no user "'.$search.'" found in our database :(</div>';
            $statement->close();
            $conn->close();

        }
        else {
            $statement->bind_result($name); # There is a error i'm encountering when I try to Select * from the line 8.
            while ($statement->fetch()) //outputs the records
            {
                echo "<div id='item'><a href=\"../user/username.php?username=$name\">$name</a></div>";
                # It supposed to show more information about the user, by using $name['first_name'] or $name['last_name']
            };
            $statement->close();
            $conn->close();
        };
    };
?>

Answer 1

您遇到的问题是mysqli_stmt::bind_result 将尝试将结果集中的每一列绑定到一个变量。这意味着您需要与列相同数量的变量。如果返回了两列，则需要将它们绑定到两个变量。

在$statement->bind_result($name); 中，您说“只有一列，所以将其绑定到$name” 而您的查询 (SELECT * FROM users WHERE username like ? LIMIT 1) 正在获取所有列那张桌子。

因此，解决方案是在这种情况下只选择您想要的单数列。替换

SELECT name 

与

SELECT *