apachebench symfony2登录页面基准
apachebench symfony2登录页面基准
提问于 2015-05-22 09:35:48
我正在尝试对symfony应用程序进行基准测试，因此我还需要对仅限于登录用户的页面进行基准测试。
我希望使用apachebench工具对应用程序进行基准测试，因此编写了一个小的shell脚本，尝试使用curl登录，获取从请求返回的phpsessid，并在apachebench命令中将其设置为cookie。
下面是looks脚本的样子：
#!/bin/bash
COOKIE_JAR="/var/www/apachebench/test.jar"
curl -c $COOKIE_JAR --data "_email=admin%40dummy.at&_password=test&_target_path=%2Fbackend" http://symfony.local/login
PHPSESSID=$(cat $COOKIE_JAR | grep PHPSESSID | cut -f 7)
ab -n 10 -p /var/www/apachebench/albumpostfile.txt -T application/x-www-form-urlencoded -C PHPSESSID=$PHPSESSID -k http://symfony.local/album/addapachebench命令应该发布表单并将数据存储在数据库中。但是，看起来我没有登录，因为数据没有存储，在使用我从浏览器复制的PHPSESSID之前，我尝试了这个命令，它运行得非常好。我也已经在全球范围内禁用了csrf的保护。
我还检查了从curl返回的PHPSESSID是否正确地输入到ab命令中。
我完全不知道我做错了什么，因为我试图通过铬扩展“邮递员”将完全相同的数据发布到登录页面，并且在那里工作。
curl请求中的cookie-jar文件如下所示：
# Netscape HTTP Cookie File
# http://curl.haxx.se/docs/http-cookies.html
# This file was generated by libcurl! Edit at your own risk.
symfony.local FALSE / FALSE 0 PHPSESSID t2glc67hlf6lrlik2ieg9r7rv7提前谢谢。
编辑：，这是apachebench在向命令中添加-v 3时的输出
WARNING: Response code not 2xx (302)
LOG: header received:
HTTP/1.0 302 Found
Date: Fri, 22 May 2015 12:03:32 GMT
Server: Apache/2.4.7 (Ubuntu)
X-Powered-By: PHP/5.5.9-1ubuntu4.9
Cache-Control: private, must-revalidate
Location: http://symfony.local/login
pragma: no-cache
expires: -1
Connection: close
Content-Type: text/html; charset=UTF-8
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8" />
<meta http-equiv="refresh" content="1;url=http://symfony.local/login" />
<title>Redirecting to http://symfony.local/login</title>
</head>
<body>
Redirecting to <a href="http://symfony.local/login">http://symfony.local/login</a>.
</body>
</html>
WARNING: Response code not 2xx (302)
LOG: header received:
HTTP/1.0 302 Found
Date: Fri, 22 May 2015 12:03:32 GMT
Server: Apache/2.4.7 (Ubuntu)
X-Powered-By: PHP/5.5.9-1ubuntu4.9
Cache-Control: private, must-revalidate
Location: http://symfony.local/login
pragma: no-cache
expires: -1
Connection: close
Content-Type: text/html; charset=UTF-8EDIT2:这是我的新shell脚本，我修改了它，使curl命令发送一个PHPSESSID及其请求。下面您可以看到两个脚本的不同输出。它看起来像第二个工作，因为它指出正确的url在重定向到Url部件。但是这一次apachebench命令根本没有做任何事情，它只是卡住了。
#!/bin/bash
COOKIE_JAR="/var/www/apachebench/test.jar"
#curl -c $COOKIE_JAR -v -d "_email=admin%40dummy.at&_password=test&_target_path=%2Fbackend" -b "PHPSESSID=1hrfrnud407n5j42oki13655g7" http://symfony.local/login
curl -c $COOKIE_JAR -v -d "_email=admin%40dummy.at&_password=test&_target_path=%2Fbackend" http://symfony.local/login
PHPSESSID=$(cat $COOKIE_JAR | grep PHPSESSID | cut -f 7)
ab -n 10 -p /var/www/apachebench/albumpostfile.txt -T application/x-www-form-urlencoded -C PHPSESSID=$PHPSESSID http://symfony.local/album/addOLD-CURL-OUTPUT:
* Hostname was NOT found in DNS cache
* Trying 127.0.1.1...
* Connected to symfony.local (127.0.1.1) port 80 (#0)
> POST /login HTTP/1.1
> User-Agent: curl/7.35.0
> Host: symfony.local
> Accept: */*
> Content-Length: 62
> Content-Type: application/x-www-form-urlencoded
>
* upload completely sent off: 62 out of 62 bytes
< HTTP/1.1 302 Found
< Date: Fri, 22 May 2015 12:39:05 GMT
* Server Apache/2.4.7 (Ubuntu) is not blacklisted
< Server: Apache/2.4.7 (Ubuntu)
< X-Powered-By: PHP/5.5.9-1ubuntu4.9
* Added cookie PHPSESSID="l2pfvtum211bd8tnpp1i0vpcj1" for domain symfony.local, path /, expire 0
< Set-Cookie: PHPSESSID=l2pfvtum211bd8tnpp1i0vpcj1; path=/
< Cache-Control: no-cache
< Location: http://symfony.local/login
< Transfer-Encoding: chunked
< Content-Type: text/html; charset=UTF-8
<
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8" />
<meta http-equiv="refresh" content="1;url=http://symfony.local/login" />
<title>Redirecting to http://symfony.local/login</title>
</head>
<body>
Redirecting to <a href="http://symfony.local/login">http://symfony.local/login</a>.
</body>
* Connection #0 to host symfony.local left intact
</html>NEW-CURL-OUTPUT:
* Hostname was NOT found in DNS cache
* Trying 127.0.1.1...
* Connected to symfony.local (127.0.1.1) port 80 (#0)
> POST /login HTTP/1.1
> User-Agent: curl/7.35.0
> Host: symfony.local
> Accept: */*
> Cookie: PHPSESSID=1hrfrnud407n5j42oki13655g7
> Content-Length: 62
> Content-Type: application/x-www-form-urlencoded
>
* upload completely sent off: 62 out of 62 bytes
< HTTP/1.1 302 Found
< Date: Fri, 22 May 2015 12:40:07 GMT
* Server Apache/2.4.7 (Ubuntu) is not blacklisted
< Server: Apache/2.4.7 (Ubuntu)
< X-Powered-By: PHP/5.5.9-1ubuntu4.9
* Added cookie PHPSESSID="3ehl5ldkbd4ngl2er663899km1" for domain symfony.local, path /, expire 0
< Set-Cookie: PHPSESSID=3ehl5ldkbd4ngl2er663899km1; path=/
< Cache-Control: no-cache
< Location: http://symfony.local/backend
< Transfer-Encoding: chunked
< Content-Type: text/html; charset=UTF-8
<
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8" />
<meta http-equiv="refresh" content="1;url=http://symfony.local/backend" />
<title>Redirecting to http://symfony.local/backend</title>
</head>
<body>
Redirecting to <a href="http://symfony.local/backend">http://symfony.local/backend</a>.
</body>
* Connection #0 to host symfony.local left intact
</html>This is ApacheBench, Version 2.3 <$Revision: 1528965 $>
Copyright 1996 Adam Twiss, Zeus Technology Ltd, http://www.zeustech.net/
Licensed to The Apache Software Foundation, http://www.apache.org/
Benchmarking symfony.local (be patient)...输出的最后一行是ab命令被卡住的地方。
回答 1
Stack Overflow用户
发布于 2015-05-23 11:51:05
我想出来了，现在起作用了。
第二个命令非常好，所以如果您对symfony进行基准测试，那么您确实需要将一个PHPSESSID传递给登录调用。
它挂在apachebench命令上的原因与shell脚本无关，而是因为代码中缺少一个{，我在调试它时不小心删除了它。通过查看apache2 error.log文件，我能够找到它。
因此，如果其他任何人都会遇到同样的问题，请记住在curl命令中添加一个PHPSESSID，登录就会正常工作，并且您可以对需要登录的页面进行基准测试。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/30393255
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