PHP致命错误:调用时传递引用已被删除 [英] PHP Fatal error: Call-time pass-by-reference has been removed

查看:110
本文介绍了PHP致命错误:调用时传递引用已被删除的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我的脚本很旧,最近又收到此错误:

I have an oldish script and lately I get this error:

Fatal error: Call-time pass-by-reference has been removed in /****/******/public_html/****/cp-list-summary.php on line 100

在该文件的第100行周围看起来像这样:

And it looks like this around line 100 on that file:

if ($row[images])
{
    $image_set = array ();
    $result = mysql_query ('SELECT fname FROM ' . $dbimgs . ' WHERE listid=\'' . $_GET['id'] . '\' ORDER BY id ASC', $link);
    while ($images = mysql_fetch_array ($result))
    {
        array_push (&$image_set, $images[fname]);
    }
}

什么原因导致该错误以及如何解决?我不是开发人员,所以请慢一点.

What causes the error and how to fix it? I'm not a developer, so please take it slow.

推荐答案

您正在尝试将指针传递给array_push中的数组.这就是为什么会遇到致命错误的原因.只需使用:

You are trying to pass a pointer to your array in array_push. That is why the fatal error is encountered. Simply use:

array_push( $image_set, $images[fname] );

注意:如果第一个参数不为 array_push() ,则会发出警告 数组.

Note: array_push() will raise a warning if the first argument is not an array.

这篇关于PHP致命错误:调用时传递引用已被删除的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆