Question

我正在寻找可以从Mac OS X终端运行的单线程的帮助。我在Mac上使用MAMP进行Web开发。我的“/ Applications / MAMP / htdocs”目录中有很多CakePHP项目。为简单起见，我只想说我有两个CakePHP项目，这是find /Applications/MAMP/htdocs -type d -iname Controller*命令的输出：

/Applications/MAMP/htdocs/my_cake1.3_project/app/controllers
/Applications/MAMP/htdocs/my_cake1.3_project/app/tests/cases/controllers
/Applications/MAMP/htdocs/my_cake1.3_project/cake/console/templates/skel/controllers
/Applications/MAMP/htdocs/my_cake1.3_project/cake/console/templates/skel/tests/cases/controllers
/Applications/MAMP/htdocs/my_cake1.3_project/cake/libs/controller
/Applications/MAMP/htdocs/my_cake1.3_project/cake/tests/cases/libs/controller
/Applications/MAMP/htdocs/my_cake1.3_project/cake/tests/test_app/controllers
/Applications/MAMP/htdocs/my_cake1.3_project/cake/tests/test_app/plugins/test_plugin/controllers
/Applications/MAMP/htdocs/my_cake2_project/app/Controller
/Applications/MAMP/htdocs/my_cake2_project/app/Test/Case/Controller
/Applications/MAMP/htdocs/my_cake2_project/lib/Cake/Console/Templates/skel/Controller
/Applications/MAMP/htdocs/my_cake2_project/lib/Cake/Console/Templates/skel/Test/Case/Controller
/Applications/MAMP/htdocs/my_cake2_project/lib/Cake/Controller
/Applications/MAMP/htdocs/my_cake2_project/lib/Cake/Test/Case/Controller
/Applications/MAMP/htdocs/my_cake2_project/lib/Cake/Test/test_app/Controller
/Applications/MAMP/htdocs/my_cake2_project/lib/Cake/Test/test_app/Plugin/TestPlugin/Controller

现在，有时候我想找到一段代码，我知道我在我的一个CakePHP项目的控制器中使用过，但我不记得它是哪个项目，所以我想搜索所有这些代码。我不想浪费时间在“app / tests / cases / controllers”文件夹或“cake /”中的任何文件夹中搜索。 find /Applications/MAMP/htdocs -type d -iname Controller* | grep -i /app/Controller命令为我提供了我要搜索的文件夹列表：

/Applications/MAMP/htdocs/my_cake1.3_project/app/controllers
/Applications/MAMP/htdocs/my_cake2_project/app/Controller

我只需要找到一种方法来获取该输出，在每行的末尾添加斜杠和星号（/ *），并将每一行传递给grep -il "string to search for"命令。任何帮助，将不胜感激。谢谢！

Answer 1

你试过这个吗？

find /Applications/MAMP/htdocs -type d -iname Controller* 
-exec grep -il "string to search for" {} /;

Answer 2

解决方案1 

也许你想检查一下find命令的两个选项：(i)path和regex 使用它们，您可以缩小查找结果并将找到的文件传递到grep -il "searchString"，例如|xargs。它看起来像：

    find /Applications/MAMP/htdocs -type f -ipath "*/app/Controller/*.php" 
| xargs grep -il 'foo'

使用-regex的

会更灵活。

解决方案2

但是如果你真的想要：

找到一种方法来获取该输出，添加斜杠和星号（/ *）每行的结尾，并将每一行传递给grep -il“要搜索的字符串为“命令。

（顺便说一句，这里“管道”不起作用。）

你可以这样做：

find .(your original find).. |grep -i "/app/Controller" 
 |sed -r 's#^(.*)$#grep -il "foo" \1/*#g'|sh

诀窍是由sed....|sh完成的。 sed行将选择前一个grep的结果，添加grep命令和选项：（grep -il "foo"）并附加“/*”以构造完整的grep命令。最后管道sh，执行它。

协助“查找”和“grep”命令

2 个答案: