如何从目录中递归找到的所有文件中搜索行匹配

Using linux commands, is there any way to know which file has the given line of code from a given project directory?

使用linux命令,有没有办法知道哪个文件具有给定项目目录中的给定代码行?

All directories are traced recursively for all files in subdirectories as well.

对于子目录中的所有文件,所有目录都以递归方式进行跟踪。

say your project is in php and is in directory /var/www/projectDir/

说你的项目在php中,位于目录/ var / www / projectDir /

and the line to be searched is public function getProjectData( to know where (in which file lying in which subdirectory ) getProjectData php function is defined?

并且要搜索的行是公共函数getProjectData(要知道getProjectData php函数定义在哪里(在哪个文件中位于哪个子目录中)?

3 个解决方案

#1

You can use the shell :

你可以使用shell:

 grep "text string to search" directory-path -R

EDIT: adding -R for recursive search

编辑:添加-R进行递归搜索

or even better, use PHP :

甚至更好,使用PHP:

$reflFunc = new ReflectionFunction('function_name');
print 'File :' $reflFunc->getFileName() . ' at line : ' . $reflFunc->getStartLine();`

#2

use the command

使用该命令

 find /var/www/projectDir/ -name '*.php'|xargs grep 'public function getProjectData('

the first portion

第一部分

find /var/www/projectDir/ -name '*.php'

will search all the files which are having extension php and within directory /var/www/projectDir

将搜索所有具有扩展名php的文件,并在目录/ var / www / projectDir中搜索

the second portion

第二部分

xargs grep 'public function getProjectData('

will search all the files found in first portion for public function getProjectData(.

将搜索第一部分中找到的所有文件,用于公共函数getProjectData(。

xargs is used to consider the output of first portion as a standard input for second portion.

xargs用于将第一部分的输出视为第二部分的标准输入。

The symbol | named pipe will pipe the output to second portion

符号|命名管道将输出管道传输到第二部分

output

/var/www/projectDir/sub/directory/somephpfile.php:    public function getProjectData($param1, $param2) {

now you can open the file and search for which line the content is defined using

现在,您可以打开文件并搜索使用的内容定义的行

(ctrl + F for gedit) or ( ctrl + W for nano)

or use any of your favorite editor.

或使用您喜欢的任何编辑器。

#3

I go with the below line. It helps me always,

我跟下面这行。它总能帮到我,

find dirname -type f -print | xargs grep -i "pattern" > /tmp/samplefile

#1