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HomeBackend DevelopmentPHP TutorialPHP客户端socket连接EXE服务端的相关问题

PHP运行环境:Windows Server 2012 R2 VL (Aliyun - QingDao) + PHP5.4.44 + MYSQL 5.6 + IIS8.5 + URL Rewrite 2.0 + FastCGI
服务端EXE运行环境:Windows Server 2012 R2 VL (Aliyun - QingDao) + E语言 5.3

function send($in){	try{		$socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);		socket_set_option($socket, SOL_SOCKET, SO_RCVTIMEO, array("sec" => 5, "usec" => 0));		socket_set_option($socket, SOL_SOCKET, SO_SNDTIMEO, array("sec" => 180, "usec" => 0));		$result = socket_connect($socket, '127.0.0.1', 9999);		$out = "";		socket_write($socket, $in, strlen($in));		$out = socket_read($socket, 10240);		socket_close($socket);		return $out;	}catch(Exception $e){		exit ("<script>alert('ERR::0');</script>");	}}

这是我的代码,正常使用,之前没加trycatch语句,就会导致,如果一个访问者打开这个页面,另一个访问者同时也打开了这个页面,那第二个用户就会报500服务器错误,经过一些大神帖子发现是因为链接服务器失败导致的,可是我的服务端EXE不限制连接数,而且第一个能连接第二个怎么就不能了呢?难道一个PHP-CGI.EXE进程在同时只能发起一个Socket连接吗?
补充:第一个用户打开网页已发起Socket 但未处理完毕 未关闭连接,第二个用户就访问该页并创建socket,则第二个用户连接失败报错。
所有的错误都是提示在socket_connect语句的所在行。


回复讨论(解决方案)

你作为服务端的程序是如何启动的?

既然你已经有了异常处理,那就不应该隐藏掉异常信息
出错时也应给出错误信息
这样判断错误原因才能有依据,你说对吧?

你echo下你的异常信息,或者使用socket_strerror(socket_last_error($socket))输出下socket产生的错误

服务器端的程序是exe 啊 当然就直接双击啊,我没有做成系统服务。。。PHP返回的异常信息就是unexpected error on socket_connect at line 5。连接失败,都unexpected了。。。socket_strerror(socket_last_error($socket))应该放在哪?

你作为服务端的程序是如何启动的?

既然你已经有了异常处理,那就不应该隐藏掉异常信息
出错时也应给出错误信息
这样判断错误原因才能有依据,你说对吧?

服务器端的程序是exe直接双击启动的 没有写系统服务项。另外我之前没做异常处理,这是后来加的,无视就好。
function send($in){        $socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);        socket_set_option($socket, SOL_SOCKET, SO_RCVTIMEO, array("sec" => 5, "usec" => 0));        socket_set_option($socket, SOL_SOCKET, SO_SNDTIMEO, array("sec" => 180, "usec" => 0));        $result = socket_connect($socket, '127.0.0.1', 9999);        $out = "";        socket_write($socket, $in, strlen($in));        $out = socket_read($socket, 10240);        socket_close($socket);        return $out;}

你echo下你的异常信息,或者使用socket_strerror(socket_last_error($socket))输出下socket产生的错误

PHP返回的异常信息就是unexpected error on socket_connect at line 5


你echo下你的异常信息,或者使用socket_strerror(socket_last_error($socket))输出下socket产生的错误

PHP返回的异常信息就是unexpected error on socket_connect at line 5
if ($result == false)         echo socket_strerror(socket_last_error($socket));

if ($result == false)         echo socket_strerror(socket_last_error($socket));


恩不用了,已经解决了。谢谢。


if ($result == false)         echo socket_strerror(socket_last_error($socket));


恩不用了,已经解决了。谢谢。 分享下经验,大家学习下
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