Question

我见过这个问题：Regular expression to match a line that doesn't contain a word?

但我无法让它发挥作用。我有一个shell脚本，我正在使用

string1.*string2.*string3

按顺序搜索文件中的3个单词。但我想改变它，以便如果badword5在该文件中的那些单词之间的任何位置，则没有与grep的正则表达式匹配。

所以这应该匹配：

./testing/test.txt:   let prep = "select string1, dog from cat",
          " where apple = 1",
          " and string2 = 2",
          " and grass = 8",
          " and string3 = ?"

但这不应该：

   ./testing/test.txt:   let prep = "select string1, dog from cat",
          " where apple = 1",
          " and string2 = 2",
          " and grass = 8",
          " and badword5 = 4", 
          " and string3 = ?"

我没试成功：

string1((?!badword5)|.)*string2((?!badword5)|.)*string3

整个脚本：

find . -name "$file_to_check" 2>/null | while read $FILE
do
   tr '\n' ' ' <"$FILE" | if grep -q "string1.*string2.*string3"; then echo "$FILE" ; fi
done >> $grep_out

Answer 1

“按顺序在文件中搜索3个单词。但是我想更改它，以便如果badword5位于该文件中的那些单词之间的任何位置，则grep没有正则表达式匹配。“ 的

确实，搜索模式延伸了多行让我们暂时放弃grep并尝试不同的东西：

#!/bin/bash find . -name "$file_to_check" 2>/dev/null | while read FILE do SCORE=0 tr ' ' '\n' <"$FILE" | while read WORD do case $WORD in "word1" ) [ $SCORE = 0 ] && SCORE=1 ;; "word2" ) [ $SCORE = 1 ] && SCORE=2 ;; "word3" ) [ $SCORE = 2 ] && echo "$FILE" && break ;; "badword5" ) SCORE=0 ;; esac done done >grep_out

案例行做了以下事情：

" word1" ) [ $SCORE = 0 ] && SCORE = 1 ;; when word1 is found: and SCORE is equal to 0 then make SCORE equal to 1 when word2 is found: and SCORE is equal to 1 then make SCORE equal to 2 when word3 is found: and SCORE is equal to 2 then print filename and break out of the inner loop.

Answer 2

您可以使用grep -v跳过badword5的行：

tr '\n' ' ' < "$FILE" | grep -v 'badword5' | if grep -q "string1.*string2.*string3"; then echo "$FILE" ; fi

如何与正则表达式进行反向匹配/黑名单？

2 个答案: