xdebug

（1）编译打包、执行springboot项目 本地：mvn clean package -U --settings D:\devsoft\apache-maven-3.5.3\conf\hubs1settings.xml -Dmaven.test.skip=true -Pdev 服务器：mvn clean package -U -Dmaven.test.skip=true -Ptest （2）运行springboot项目 java -jar xxx.jar //pinpoint-agent监控的方式启动springboot项目 nohup java -Xms6144m -Xmx6144m -XX:PermSize=1024m -XX:MaxPermSize=1024m -javaagent:/home/xpp/apm/pinpoint-agent/pinpoint-bootstrap.jar -Dpinpoint.agentId=PP_123.16.xx.115 -Dpinpoint.applicationName=crs_bb_bb-xxx-data -jar xxx.jar （3）SpringBoot开启远程debug与远程监控 a.开启远程debug nohup java -Xdebug -Xrunjdwp:server=y,transport=dt_socket

I have installed xdebug, I can see in phpinfo() that it's installed (but it's OFF). However, I don't want to enable it for the whole server/apache2, I just want to enable it for one virtual host. How can I do this? You can set xdebug in php.ini with off value: zend_extension=/usr/lib/php/modules/xdebug.so xdebug.remote_enable off xdebug.remote_port 9000 xdebug.idekey PHP-XDEBUG and turn on only .htaccess with directive: php_flag xdebug.remote_enable on 来源： https://stackoverflow.com/questions/15423705/how-can-i-use-xdebug-to-debug-only-one-virtual-host

安装xdebug 下载xdebug 可到 xdebug官网 下载与自己PHP版本对应的xdebug扩展 安装 [root] # cd /usr/local/src [root] # tar zxvf xdebug-2.6.0RC2.tgz [root] # cd xdebug-2.6.0RC2 [root] # /usr/local/php/bin/phpize [root] # ./configure --enable-xdebug --with-php-config=/usr/local/php/bin/php-config [root] # make && make install 配置xdebug 修改php.ini配置文件，在文件中追加以下内容 zend_extension=xdebug.so //指定Xdebug扩展文件的路径 xdebug.remote_enable=1 //是否开启远程调试 xdebug.remote_handler=dbgp //指定远程调试的处理协议 xdebug.remote_mode=req //可以设为req或jit，req表示脚本一开始运行就连接远程客户端，jit表示脚本出错时才连接远程客户端。 xdebug.remote_host=192.168.1.98 //指定远程调试的主机名（安装phpstorm的主机ip） xdebug

I am running a PHP daemon to be profiled. The started php process loads all required data, forks itself to distribute workload over all cores, waits for the forked children to finish, and collects the results generated by the children. Since I am sharing CLI environment with other users, I need to start xdebug profiling by injecting php.ini values into the shell call. $ php -d xdebug.profiler_enable=1 -d xdebug.profiler_output_dir="/home/xxx" daemon.php The generated cachegrind-file, how ever, profiles the parent and hence shows 90% sleep. Is there a way to profile the workers without building

由于以前使用的都是windows对于新买的Mac系统异常不熟悉，所以安装环境碰到了许多小问题，这里做些总结。希望可以对以后安装的朋友有点用处。 首先是搭建PHP 环境，这里我使用了下面这个博客，讲的比较详细 全新安装Mac OSX 开发者环境 同时使用homebrew搭建 PHP，Nginx ，MySQL，Redis，Memcache ... ... (LNMP开发环境) http://segmentfault.com/a/1190000000606752 但是还是碰到一些小问题； 因为我想安装PHP56 所以讲其中55的地方都需要改掉。（这点应该都很容易理解 ^_^） 然后安装了 zsh ，那么你的shell 就已经变成zsh了。这时请注意 博文中替换的环境变量是这样的 echo 'export PATH="$(brew --prefix php55)/bin:$PATH"' >> ~/.bash_profile #for php echo 'export PATH="$(brew --prefix php55)/sbin:$PATH"' >> ~/.bash_profile #for php-fpm echo 'export PATH="/usr/local/bin:/usr/local/sbib:$PATH"' >> ~/.bash_profile #for other

I setup Laravel Homestead. I then configured both homestead xdebug.ini and PHPStorm to make the debugging work. Here is my xdebug.ini inside homestead zend_extension=xdebug.so xdebug.remote_autostart = on xdebug.remote_enable = on xdebug.remote_connect_back = on xdebug.remote_port = 9000 xdebug.idekey = "vagrant" To start a debugging session the steps I follow are In PHPStorm --> Start Listening for connections In PHPStorm set a breakpoint In my browser --> Use XDebug Chrome Helper OR add to my URL ?XDEBUG_SESSION_START= Load the page This works perfectly. My problem is when I'm inside