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如何使用 grep 删除多行字符串

tag-icon • tag-icon command-line text-processing grep
如何使用 grep 删除多行字符串

假设我有file.txt以下内容（例如简化的。实际上它要复杂得多）：

This line I DON'T want to remove \
    common substring

This line I want to remove \
    common substring

我试过

mv file.txt __file.txt
grep -vwE "This line I want" __file.txt > file.txt
rm __file.txt

重新创建file.txt并删除整个'This line I want to remove common substring'与线闸，但作为输出我得到

This line I DON'T want to remove \
    common substring


    common substring

我不能grep -vwE "common substring" __file.txt > file.txt另外使用因为它会影响另一个多行字符串......

那么如何让 grep 在考虑行中断的情况下删除整个多行？

答案1

应删除分成 8 行的行

如果每个“段落”之间有一个空行，那么你可以写

awk -F '\n' -v RS= -v ORS='\n\n' 'NF != 8'

演示：

cat <<"END" | awk -F'\n' -v RS= -v ORS='\n\n' 'NF != 8'
This line I DON'T want to remove \
    common substring

This line I want to remove \
    common substring \
    3 \
    4 \
    5 \
    6 \
    7 \
    8

Last line
END

This line I DON'T want to remove \
    common substring

Last line

请注意，内置的 bashread可以遵循行连续性，但它们的位置被抹去了：

while IFS= read line; do declare -p line; done << "END"
This line I DON'T want to remove \
    common substring

This line I want to remove \
    common substring \
    3 \
    4 \
    5 \
    6 \
    7 \
    8

Last line
END

declare -- line="This line I DON'T want to remove     common substring"
declare -- line=""
declare -- line="This line I want to remove     common substring     3     4     5     6     7     8"
declare -- line=""
declare -- line="Last line"

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