bash grep - 列出所有不包含这两种模式的行
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grep - List all lines not containing both pattern
提问by Newbie
I have a text file having some records. I have two patterns to verify and I want to list all lines from the file not containing both pattern. How can I do this using grepcommand?
I tried few things using grep -vbut nothing seem to work.
我有一个包含一些记录的文本文件。我有两个模式要验证，我想列出文件中不包含这两个模式的所有行。如何使用grep命令执行此操作？
我尝试了几件事，grep -v但似乎没有任何效果。
Suppose my text file is as follows.
1. qwerpattern1yui
2. adspattern2asd
3. cczxczc
4. jkjkpattern2adsdapattern1
假设我的文本文件如下。
1. qwer pattern1yui
2. ads pattern2asd
3. cczxczc
4. pattern2jkjk adsdapattern1
I want to list lines 1, 2 and 3 only.
我只想列出第 1、2 和 3 行。
Thanks in advance.
提前致谢。
回答by anubhava
You can use:
您可以使用：
grep -w -v -e "word1" -e "word2" file
OR else using egrep:
或其他使用egrep：
egrep -w -v -e "word1|word2" file
UPDATE:Based on comments, it seems following awk will work better:
更新：根据评论，似乎遵循 awk 会更好：
awk '!(/pattern1/ && /pattern2/)' file
回答by Digital Trauma
If I'm keeping up with the comments and edits right, I think this is what you need:
如果我跟上评论和编辑的权利，我认为这就是您所需要的：
$ grep -E -v 'pattern1.*pattern2|pattern2.*pattern1' test 1. qwerpattern1yui 2. adspattern2asd 3. cczxczc $
回答by Jotne
If you like to try awk
如果你喜欢尝试 awk
awk '!/pattern1|pattern2/' file
It will not print any lines if it contains any of the patters
如果它包含任何模式，则不会打印任何行
You can also expand this:
你也可以扩展这个：
awk '!/pattern1|pattern2|pattern3|pattern4/' file
Example
例子
cat file
one
two
three
four
one two
two
nine
six two
remove all lines with oneor twoor both of them.
删除所有带有one或two或两者的行。
awk '!/one|two/' file
three
four
nine
回答by Digital Trauma
While the standard tools-based answers (awk, grep, etc) are generally simpler and more straightforward, for completion if you needed a pure-bash solution, you could do this:
虽然基于标准工具的答案（awk、grep等）通常更简单、更直接，但如果您需要纯 bash 解决方案来完成，您可以这样做：
$ while IFS= read -r ln; do [[ $ln =~ pattern1 ]] && [[ $ln =~ pattern2 ]] && continue; printf "%s\n" "$ln"; done < test
1. qwerpattern1yui
2. adspattern2asd
3. cczxczc
$
