从多个文件的grep搜索中获得最后一行 [英] get last line from grep search on multiple files

问题描述

我找到了只显示grep搜索最后一行的方法：

  grep PATERN FILE_NAME | tail -1 
  

我也找到了在多个选定文件中进行grep搜索的方法：

  find。 -nameFILE_NAME| xargs -I name grep PATERN name 
  

现在我只希望得到grep结果的最后一行为每个单个文件。
我试过这个：

  find。 -nameFILE_NAME| xargs -I名称grep PATERN名称| tail -1 
  

这只会返回我想要的最后一个文件的最后一个值最后匹配每个文件的模板。

解决方案

  文件名）;请grep PATERN $ f |尾巴-1;完成
  

I'm curently having some problem with a grep command.

I've found the way to only show the last line of a grep search :

grep PATERN FILE_NAME | tail -1

I also find the way to make a grep search in multiple selected files :

find . -name "FILE_NAME" | xargs -I name grep PATERN name

Now I would like to only get the last line of the grep result for each single file. I tried this :

 find . -name "FILE_NAME" | xargs -I name grep PATERN name | tail -1

This returns me only the last value of the last file where I would like to have the last matching patern for every file.

解决方案

for f in $(find . -name "FILE_NAME"); do grep PATERN $f | tail -1; done

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