PHP 5.4调用时间传递引用

PHP 5.4调用时间传递引用 - 这里很容易解决？(PHP 5.4 Call-time pass-by-reference - Easy fix here?)

我有这个错误消息与Centos 5.9，PHP 5.4和较旧的PHP程序扩展（typo3 CMS）。

PHP致命错误：已在第279行的class.tx_spscoutnetcalendar_pi1.php中删除了调用时传递引用

这是模拟php代码的功能：

// ********* Start XML code ********* // get XML data from an URL and return it function fetchCalendarData($xmlUrl,$timeout) { $xmlSource=""; $url = parse_url($xmlUrl); $fp = fsockopen($url['host'], "80", &$errno, &$errstr, $timeout); if ($fp) { fputs($fp, "GET ".$url['path']."?".$url['query']." HTTP/1.1\r\nHost: " . $url['host'] . "\r\n\r\n"); while(!feof($fp)) $xmlSource .= fgets($fp, 128); } // strip HTTP header if ($pos = strpos($xmlSource,"<?xml")) { // this has to be the first line $xmlSource = substr($xmlSource, $pos); } else { $xmlSource=""; } // I have no idea why, but at the end of the fetched data a '0' breaks the XML syntax and provides an // error message in the parser. So I just cut the last 5 characters of the fetched data $xmlSource = substr($xmlSource,0,strlen($xmlSource)-5); return $xmlSource; }

具体这一行279

$fp = fsockopen($url['host'], "80", &$errno, &$errstr, $timeout);

请帮忙，我不是php专家。

I have this error Message with Centos 5.9, PHP 5.4 and an older PHP program extension (typo3 CMS).

PHP Fatal error: Call-time pass-by-reference has been removed in class.tx_spscoutnetcalendar_pi1.php on line 279

This is analog the php code function:

// ********* Start XML code ********* // get XML data from an URL and return it function fetchCalendarData($xmlUrl,$timeout) { $xmlSource=""; $url = parse_url($xmlUrl); $fp = fsockopen($url['host'], "80", &$errno, &$errstr, $timeout); if ($fp) { fputs($fp, "GET ".$url['path']."?".$url['query']." HTTP/1.1\r\nHost: " . $url['host'] . "\r\n\r\n"); while(!feof($fp)) $xmlSource .= fgets($fp, 128); } // strip HTTP header if ($pos = strpos($xmlSource,"<?xml")) { // this has to be the first line $xmlSource = substr($xmlSource, $pos); } else { $xmlSource=""; } // I have no idea why, but at the end of the fetched data a '0' breaks the XML syntax and provides an // error message in the parser. So I just cut the last 5 characters of the fetched data $xmlSource = substr($xmlSource,0,strlen($xmlSource)-5); return $xmlSource; }

And specific this line 279

$fp = fsockopen($url['host'], "80", &$errno, &$errstr, $timeout);

Any help here please, i'm not a php expert.

最满意答案

只需删除前导&喜欢这个：

$fp = fsockopen($url['host'], "80", $errno, $errstr, $timeout);

变量仍然通过引用传递，但是您不需要使用&来表示从PHP 5.4开始。

Just remove the leading & like this:

$fp = fsockopen($url['host'], "80", $errno, $errstr, $timeout);

the variables are still passed by reference, but you don't need the & to indicate that from PHP 5.4 onwards.