Bash，Linux，需要根据另一个文件中的匹配内容从一个文件中删除行 [英] Bash, Linux, Need to remove lines from one file based on matching content from another file

问题描述

当同一行存在于另一个文件中时，如何删除该文件中的行有很多示例.我已经通读了它们，如果整行匹配，它们都会删除.例如:grep -vxF -f file1 file2

There are many examples on how to remove lines in one file when that same line exists in another file. I have read through them and they all remove if the full line matches. Examples like: grep -vxF -f file1 file2

我所拥有的略有不同.我有一个来自我的网站和我的客户网站的 URL 列表.当域与另一个文件中的域匹配时，我想从该文件中删除行.

What I have is slightly different. I have a list of URLs from my websites and my clients websites. I want to remove lines from that file when the domain matches a domain in another file.

所以第一个文件可能看起来像:

So the first file might look like:

http://www.site1.com/some/path
http://www.site2.com/some/path
http://www.site3.com/some/path
http://www.site4.com/some/path

第二个文件可能是:

site2.com
www.site4.com

我希望输出为:

http://www.site1.com/some/path
http://www.site3.com/some/path

推荐答案

grep 标志太多.具体来说:-x 将使您无法获得所需的结果.

You have too many grep flags. Specifically: -x will keep you from getting your desired results.

假设 file1 具有模式，而 file2 具有 URL，只需使用:

Assuming that file1 has the patterns, and file2 has the URLs, just use:

grep -v -f file1 file2

-x 标志会阻止你得到你想要的结果:使用 -x 意味着:只匹配整行，即只匹配一行，如果这条线正是，例如site2.com.

The -x flag will keep you from getting the results that you want: using -x means: match only against the entire line, i.e. only match a line if the line is exactly, e.g. site2.com.

来自man grep:

-x, --line-regexp

-x, --line-regexp

仅选择与整行完全匹配的匹配项.

Select only those matches that exactly match the whole line.

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