bash 删除包含多个字符串之一的行
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Delete line containing one of multiple strings
提问by user1899415
I have a text file and I want to remove all lines containing the words: facebook, youtube, google, amazon, dropbox, etc.
我有一个文本文件，我想删除所有包含以下单词的行：facebook, youtube, google, amazon, dropbox, 等。
I know to delete lines containing a string with sed:
我知道用 sed 删除包含字符串的行：
sed '/facebook/d' myfile.txt
I don't want to run this command five different times though for each string, is there a way to combine all the strings into one command?
我不想为每个字符串运行这个命令五次，有没有办法将所有字符串组合成一个命令？
回答by gpojd
Try this:
尝试这个：
sed '/facebook\|youtube\|google\|amazon\|dropbox/d' myfile.txt
From GNU's sed manual:
来自GNU 的 sed 手册：
regexp1\|regexp2Matches either
regexp1orregexp2. Use parentheses to use complex alternative regular expressions. The matching process tries each alternative in turn, from left to right, and the first one that succeeds is used. It is a GNU extension.
regexp1\|regexp2匹配
regexp1或regexp2。使用括号来使用复杂的替代正则表达式。匹配过程从左到右依次尝试每个选项，并使用第一个成功的选项。它是一个 GNU 扩展。
回答by glenn Hymanman
grep -vf wordsToExcludeFile myfile.txt
"wordsToExcludeFile" should contain the words you don't want, one per line.
“wordsToExcludeFile”应包含您不想要的单词，每行一个。
If you need to save the result back to the same file, then add this to the command:
如果您需要将结果保存回同一个文件，请将其添加到命令中：
> myfile.new && mv myfile.new myfile.txt
回答by jaypal singh
With awk
和 awk
awk '!/facebook|youtube|google|amazon|dropbox/' myfile.txt > filtered.txt
