Unix 命令列出包含字符串但*不*包含另一个字符串的文件
如何递归查看包含一个字符串且不包含另一个字符串的文件列表？另外，我的意思是评估文件的文本，而不是文件名。
结论：
根据评论，我最终使用：
find . -name "*.html" -exec grep -lR 'base\-maps' {} \; | xargs grep -L 'base\-maps\-bot'
这返回了带有“base-maps”而不是“base-maps-bot”的文件。谢谢你！！
How do I recursively view a list of files that has one string and specifically doesn't have another string? Also, I mean to evaluate the text of the files, not the filenames.
Conclusion:
As per comments, I ended up using:
find . -name "*.html" -exec grep -lR 'base\-maps' {} \; | xargs grep -L 'base\-maps\-bot'
This returned files with "base-maps" and not "base-maps-bot". Thank you!!
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试试这个：
解释：
grep -lr使 grep 递归地 (r) 输出包含grep -L。grep -L仅当文件不包含Try this:
Explanation:
grep -lrmakes grep recursively (r) output a list (l) of all files that contain<string-to-match>. xargs loops over these files, callinggrep -Lon each one of them.grep -Lwill only output the filename when the file does not contain<string-not-to-match>.在上面的答案中使用 xargs 是没有必要的；你可以像这样实现同样的事情：
grep -q意味着安静地运行，但返回一个退出代码，指示是否找到匹配项；然后，find可以使用该退出代码来确定是否继续执行其其余选项。如果-exec grep -q <要匹配的字符串> {} \;返回 0，然后继续执行-not -exec grep -q{} \; 。如果也返回 0，它将继续执行-print，打印文件名。正如另一个答案所指出的，以这种方式使用 find 比 grep -Rl 具有主要优势，在 grep -Rl 中您只想搜索某种类型的文件。另一方面，如果您确实想要搜索所有文件，
grep -Rl可能会更快，因为它使用一个grep进程对所有文件执行第一个过滤器，而不是为每个文件使用单独的grep进程。The use of xargs in the answers above is not necessary; you can achieve the same thing like this:
grep -qmeans run quietly but return an exit code indicating whether a match was found;findcan then use that exit code to determine whether to keep executing the rest of its options. If-exec grep -q <string-to-match> {} \;returns 0, then it will go on to execute-not -exec grep -q <string-not-to-match>{} \;. If that also returns 0, it will go on to execute-print, which prints the name of the file.As another answer has noted, using
findin this way has major advantages overgrep -Rlwhere you only want to search files of a certain type. If, on the other hand, you really want to search all files,grep -Rlis probably quicker, as it uses onegrepprocess to perform the first filter for all files, instead of a separategrepprocess for each file.这些答案似乎与 BOTH 字符串相匹配。以下命令应该效果更好：
These answers seem off as the match BOTH strings. The following command should work better:
这是一个更通用的结构：
此命令输出名称与< /code>，但不包含
find谓词），其中包含增强功能包括：
如果您不需要按名称过滤（通常希望考虑当前目录中的所有文件），您可以删除
find并将-R添加到第一个 <代码>grep：Here is a more generic construction:
This command outputs files whose name matches
<nameFilter>(adjustfindpredicates as you need) which contain<patternYes>, but do not contain<patternNo>.The enhancements are:
If you don't need to filter by name (one often wants to consider all the files in current directory), you can strip
findand add-Rto the firstgrep:寻找 。 -maxdepth 1 -name "*.py" -exec grep -L "字符串不匹配" {} \;
此命令将获取同一目录中不包含“string-not-to-match”的所有“.py”文件。
find . -maxdepth 1 -name "*.py" -exec grep -L "string-not-to-match" {} \;
This Command will get all ".py" files that don't contain "string-not-to-match" at same directory.
匹配字符串 A 并排除字符串 B 和C 出现在我使用的同一行中，并且引号允许搜索字符串包含空格
解释：grep -r 递归地过滤输出格式中匹配的所有行
从这些行中排除 (grep -v) 还包含 -e 字符串 B 或 -e 字符串 C 的行。awk 用于仅打印第一个字段（文件名），使用冒号作为字段分隔符 -F
To match string A and exclude strings B & C being present in the same line I use, and quotes to allow search string to contain a space
Explanation: grep -r recursively filters all lines matching in output format
To exclude (grep -v) from those lines the ones that also contain either -e string B or -e string C. awk is used to print only the first field (the filename) using the colon as fieldseparator -F