列出包含字符串但 *NOT* 包含另一个字符串的文件的 Unix 命令 [英] Unix Command to List files containing string but *NOT* containing another string
问题描述
如何递归查看包含一个字符串但具体没有另一个字符串的文件列表?另外,我的意思是评估文件的文本,而不是文件名.
How do I recursively view a list of files that has one string and specifically doesn't have another string? Also, I mean to evaluate the text of the files, not the filenames.
结论:
根据评论,我最终使用了:
As per comments, I ended up using:
find . -name "*.html" -exec grep -lR 'base-maps' {} ; | xargs grep -L 'base-maps-bot'
这返回了带有base-maps"而不是base-maps-bot"的文件.谢谢!!
This returned files with "base-maps" and not "base-maps-bot". Thank you!!
推荐答案
试试这个:
grep -rl <string-to-match> | xargs grep -L <string-not-to-match>
说明:grep -lr
使 grep 递归 (r) 输出包含
的所有文件的列表 (l).xargs 循环遍历这些文件,在每个文件上调用 grep -L
.grep -L
只会在文件不包含 <string-not-to-match>
时输出文件名.
Explanation: grep -lr
makes grep recursively (r) output a list (l) of all files that contain <string-to-match>
. xargs loops over these files, calling grep -L
on each one of them. grep -L
will only output the filename when the file does not contain <string-not-to-match>
.
这篇关于列出包含字符串但 *NOT* 包含另一个字符串的文件的 Unix 命令的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!