問題描述

我正在嘗試進行用戶名搜索 [我似乎已經完成并正在工作] 但是當您搜索用戶名時，會顯示有關該帳戶的信息.比如我搜索了virtualAnon，他的名字和first_name等信息就會出現在他的用戶名后面.

I am trying to do a username search [which I seemingly finished and working as well] but when you've searched for the username, the information about the account will show up. For example, I've searched for virtualAnon, his name and information such as first_name will show up after his username.

我試圖通過替換 $query = "SELECT username FROM users WHERE username like 來修復它?LIMIT 1"; to $query = "SELECT * FROM users WHERE username like ?LIMIT 1"; 但在我嘗試之后，錯誤

I've tried to fix it by replacing $query = "SELECT username FROM users WHERE username like ? LIMIT 1"; to $query = "SELECT * FROM users WHERE username like ? LIMIT 1"; but after I've tried that, the error

mysqli_stmt::bind_result(): 綁定變量數量不匹配PHP中準備好的語句中的字段數

mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in PHP

出現.

這是用于獲取用戶名和數據庫的 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();
        };
    };
?>

推薦答案

您遇到的問題是 mysqli_stmt::bind_result 將嘗試將結果集中的每一列綁定到一個變量.這意味著您需要與列相同數量的變量.如果有兩列返回，則需要將它們綁定到兩個變量.

The problem that you're getting is that mysqli_stmt::bind_result will try to bind each column in the result set to a variable. Which means that you need the same amount of variables as you've got columns. If you've got two columns being returned, you need to bind them to two variables.

在$statement->bind_result($name);中，你是說只有一列，所以將它綁定到$name" 而您的查詢(SELECT * FROM users WHERE username like ? LIMIT 1)正在獲取該表的所有列.

In $statement->bind_result($name);, you're saying "There's only going to be one column, so bind it to $name" whereas your query (SELECT * FROM users WHERE username like ? LIMIT 1) is fetching all the columns for that table.

所以解決方案是在這個實例中只選擇你想要的單數列.替換

So the solution is to only select the singular column you want in this instance. Replace

SELECT name 

與

SELECT *

這篇關于mysqli_stmt::bind_result(): 綁定變量的數量與 PHP 中準備好的語句中的字段數量不匹配的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！