注意：未定义的变量图片，性别

Question

我收到了这个错误：

Notice: Undefined variable: picture in C:\Users\Raj\PhpstormProjects\comment03\Wall\wall-functions.php on line 36

Notice: Undefined variable: picture in C:\Users\Raj\PhpstormProjects\comment03\Wall\wall-functions.php on line 38

Notice: Undefined variable: gender in C:\Users\Raj\PhpstormProjects\comment03\Wall\wall-functions.php on line 40

我已经从现在下载了一个php墙脚本，因为我连接数据库后运行程序它给了我这个错误..请帮助我现在应该做什么？

这是我的wall-functions.php代码：

<?php
$path = 'http://yepinol.com/wall/';

if(!function_exists('getUserImg')) {

        function getUserImg($user_id = ''){

            $username_get = mysql_query("SELECT picture,gender from member where member_id=".$user_id." order by member_id desc limit 1");
            while ($name = @mysql_fetch_array($username_get))
            {
                $picture = $name['picture'];
                $gender = $name['gender'];
            }

            $imageUser = 'pics/'.$picture;

            if (!file_exists($imageUser) || $picture=='')  
            {
                if($gender == 'm')
                $imageUser = 'pics/no-image-m.png';
                else
                $imageUser = 'pics/no-image-f.png';
            }

            return  $imageUser;
        }
    }
    ?>

Answer 1

尝试在while循环之前/之前声明$picture和$gender。

Answer 2

尝试以下代码

<?php
$path = 'http://yepinol.com/wall/';
$picture =''; // Declare variable
$gender = ''; // Declare variable

if(!function_exists('getUserImg')) {

        function getUserImg($user_id = ''){

            $username_get = mysql_query("SELECT picture,gender from member where member_id=".$user_id." order by member_id desc limit 1");
            while ($name = @mysql_fetch_array($username_get))
            {
                $picture = $name['picture'];
                $gender = $name['gender'];
            }

            $imageUser = 'pics/'.$picture;

            if (!file_exists($imageUser) || $picture=='')  
            {
                if($gender == 'm')
                $imageUser = 'pics/no-image-m.png';
                else
                $imageUser = 'pics/no-image-f.png';
            }

            return  $imageUser;
        }
    }
    ?>

我认为您遇到Scope of variable

的问题