bash sed- 删除不包含模式的行
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sed- delete line that doesn't contain a pattern
提问by buydadip
I'm surprised that I can't find a question similar to this one on SO.
我很惊讶我在 SO 上找不到与此类似的问题。
How do I use sed to delete all lines that do not contain a specific pattern.
如何使用 sed 删除所有不包含特定模式的行。
For example, I have this file:
例如，我有这个文件：
cat kitty dog
giraffe panda
lion tiger
I want a sed command that, when called, will delete all lines that do not contain the word cat:
我想要一个 sed 命令，当调用它时，将删除所有不包含单词的行cat：
cat kitty dog
回答by Amit Verma
This will do:
这将：
sed -i '/cat/!d' file1.txt
To force an exact match:
强制精确匹配：
sed -i '/\<cat\>/!d' file1.txt
or
或者
sed -i '/\bcat\b/!d' file1.txt
where \<\>& \b\bforce an exact match.
where \<\>&\b\b强制精确匹配。
回答by Kent
So your requirement would be "give me all lines containing string cat". then why not just simply using grep:
所以你的要求是“给我所有包含字符串的行cat”。那么为什么不简单地使用grep：
grep cat file
回答by Jotne
You can use this awk
你可以用这个 awk
awk '/cat/' file
回答by Denio Mariz
to see all lines containg word 'cat' (as pointed by Kent):
查看包含单词“cat”的所有行（如 Kent 所指出的）：
grep cat file
to see all lines NOT containg word 'cat':
查看所有不包含单词“cat”的行：
grep -v cat file
