自定义PHP函数出错:未返回任何变量 [英] Error with custom PHP function: no variable being returned

问题描述

好的，下面是我的PHP函数.我可以确认我已连接到数据库，因为可以直接在文件中使用mysqli_query函数对其进行更新.

Alright, so here is a look at my PHP function. I can confirm I AM connected to the database, as I can make updates to it with mysqli_query functions directly in the file.

<?php
function username_from_id($id) {
$id = mysqli_real_escape_string($id);
$query = mysqli_query($d,"SELECT `username` FROM `users` WHERE `id` = '$id'");
$result = mysqli_fetch_array($query);
$res = $result['username'];
return $res;
}
?>

该函数的目的是，如果用户的ID等于查询中输入的用户名，则选择该用户名，然后将其返回.在文件中，看起来像这样

The purpose of the function is to select the username of a user if their ID equals what is put into the query, then return it. In the file, it looks like this

<?php
include 'file_where_function_is.php';
$id = '1';
echo username_from_id($id);
?>

没有任何显示.有什么想法吗?

Nothing shows up. Any ideas?

推荐答案

正如评论中指出的那样，这是一个范围界定问题.您的$d变量(mysqli实例)不在username_from_id的范围内.这是解决方法...

As pointed out in comments, this is a scoping issue. Your $d variable (mysqli instance) is not in scope within username_from_id. Here's how to fix it...

function username_from_id(mysqli $d, $id) {
    if (!$stmt = $d->prepare('SELECT username FROM users WHERE id = ? LIMIT 1')) {
        throw new Exception($d->error, $d->errno);
    }
    $stmt->bind_param('i', $id);

    if (!$stmt->execute()) {
        throw new Exception($stmt->error, $stmt->errno);
    }

    $stmt->bind_result($username);

    if ($stmt->fetch()) {
        return $username;
    }
    return null;
}

并这样称呼它

include 'file_where_function_is.php';
$id = 1;
echo username_from_id($d, $id); // assuming $d exists in this scope

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