仅在递归差异中包含与给定模式匹配的文件 [英] Only include files that match a given pattern in a recursive diff

问题描述

如何对两个目录（a和b）中的文件执行递归diff：

How can you perform a recursive diff of the files in two directories (a and b):

$ diff -r a b

，但仅查看名称与给定模式匹配的文件。例如，使用find命令中可用的相同语法，这看起来像：

but only look at files whose name matches a given pattern. For example, using the same syntax available in the find command, this would look like:

$ diff -r a b -name "*crazy*"

这将显示a和b中具有相同名称和路径的文件之间的差异。

which would show diffs between files with the same name and path in a and b, which have "crazy" in their name.

有效地，我正在寻找diff中可用的--exclude选项的相反内容。

Effectively, I'm looking for the opposite of the --exclude option which is available in diff.

推荐答案

也许这是间接的，但它应该可以工作。您可以使用查找获取不匹配模式的文件列表，然后排除所有这些文件：

Perhaps this is a bit indirect, but it ought to work. You can use find to get a list of files that don't match the pattern, and then "exclude" all those files:

find a b -type f ! -name 'crazy' -printf '%f\n' | diff -r a b -X -

-X-将使 diff 从stdin读取模式，并排除任何匹配的内容。如果您的文件名称中没有 * 或？之类的有趣字符，这应该可以工作。唯一的缺点是您的差异将不包含 find 命令，因此列出的 diff 命令不是那么有用。

The -X - will make diff read the patterns from stdin and exclude anything that matches. This should work provided your files don't have funny chars like * or ? in their names. The only downside is that your diff won't include the find command, so the listed diff command is not that useful.

（我只在GNU find 和 diff ）。

(I've only tested it with GNU find and diff).

编辑：

由于仅非GNU 查找没有 -printf ， sed 可能是用作替代：

Since only non-GNU find doesn't have -printf, sed could be used as an alternative:

find a b -type f ! -name '*crazy*' -print | sed -e 's|.*/||' | diff -X - -r a b

这还假设非GNU diff 有 -X 我不知道。

That's also assuming that non-GNU diff has -X which I don't know.

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