bash Linux：删除不包含所有指定单词的文件
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Linux: Removing files that don't contain all the words specified
提问by Daniel
Inside a directory, how can I delete files that lack any of the words specified, so that only files that contain ALL the words are left? I tried to write a simple bash shell script using grep and rm commands, but I got lost. I am totally new to Linux, any help would be appreciated
在目录中，如何删除缺少任何指定单词的文件，以便只留下包含所有单词的文件？我尝试使用 grep 和 rm 命令编写一个简单的 bash shell 脚本，但我迷路了。我对 Linux 完全陌生，任何帮助将不胜感激
回答by toolkit
How about:
怎么样：
grep -L foo *.txt | xargs rm
grep -L bar *.txt | xargs rm
If a file does notcontain foo, then the first line will remove it.
如果文件确实不包含foo，则第一行会删除它。
If a file does notcontain bar, then the second line will remove it.
如果文件确实不包含bar，则第二线将其删除。
Only files containing both fooand barshould be left
只应保留包含foo和的文件bar
-L, --files-without-match
Suppress normal output; instead print the name of each input
file from which no output would normally have been printed. The
scanning will stop on the first match.
See also @Mykola Golubyev's postfor placing in a loop.
另请参阅@Mykola Golubyev 的帖子以放置在循环中。
回答by Mykola Golubyev
list=`Word1 Word2 Word3 Word4 Word5`
for word in $list
grep -L $word *.txt | xargs rm
done
回答by soulmerge
Addition to the answers above: Use the newline character as delimiter to handle file names with spaces!
除了上面的答案：使用换行符作为分隔符来处理带空格的文件名！
grep -L $word $file | xargs -d '\n' rm
回答by user65636
grep -L word | xargs rm
grep -L 字 | xargs rm
回答by Andy
To do the same matching filenames (not the contents of files as most of the solutions above) you can use the following:
要执行相同的匹配文件名（不是上述大多数解决方案的文件内容），您可以使用以下命令：
for file in `ls --color=never | grep -ve "\(foo\|bar\)"`
do
rm $file
done
As per comments:
根据评论：
for file in `ls`
shouldn't be used. The below does the same thing without using the ls
不应该使用。下面做同样的事情而不使用ls
for file in *
do
if [ x`echo $file | grep -ve "\(test1\|test3\)"` == x ]; then
rm $file
fi
done
The -ve reverses the search for the regexp pattern for either foo or bar in the filename. Any further words to be added to the list need to be separated by \| e.g. one\|two\|three
-ve 反转搜索文件名中 foo 或 bar 的正则表达式模式。要添加到列表中的任何其他单词都需要用 \| 分隔 例如一\|二\|三
回答by Dimitre Radoulov
You could try something like this but it may break if the patterns contain shellor grepmeta characters:
您可以尝试这样的操作，但如果模式包含shell或grep元字符，它可能会中断：
(in this example one two threeare the patterns)
（在这个例子中，一二三是模式）
for f in *; do
unset cmd
for p in one two three; do
cmd="fgrep \"$p\" \"$f\" && $cmd"
done
eval "$cmd" >/dev/null || rm "$f"
done
回答by paxdiablo
First, remove the file-list:
首先，删除文件列表：
rm flist
Then, for each of the words, add the file to the filelist if it contains that word:
然后，对于每个单词，如果文件包含该单词，则将文件添加到文件列表中：
grep -l WORD * >>flist
Then sort, uniqify and get a count:
然后排序，统一并得到一个计数：
sort flist | uniq -c >flist_with_count
All those files in flsit_with_count that don't have the number of words should be deleted. The format will be:
应该删除 flsit_with_count 中没有单词数的所有文件。格式将是：
2 file1
7 file2
8 file3
8 file4
If there were 8 words, then file1 and file2 should be deleted. I'll leave the writing/testing of the script to you.
如果有 8 个字，则应删除 file1 和 file2。我会把脚本的编写/测试留给你。
Okay, you convinced me, here's my script:
好的，你说服了我，这是我的脚本：
#!/bin/bash
rm -rf flist
for word in fopen fclose main ; do
grep -l ${word} *.c >>flist
done
rm $(sort flist | uniq -c | awk ' != 3 {print } {}')
This removes the files in the directory that didn't have all three words:
这将删除目录中没有所有三个单词的文件：
回答by Eugene Morozov
This will remove all files that doesn't contain words Pingor Sent
这将删除所有不包含Ping或Sent字样的文件
grep -L 'Ping\|Sent' * | xargs rm
