列出一个单词消失的文件
因此，假设我有两个分支old_release（包含旧版本）和new_release（包含新版本）。在旧版本中，有一个称为cook的工具，我认为它仍在新版本中，但已更名。我需要弄清楚它的重命名。
我想要一个必须满足标准的文件列表；
- 文件都存在于
old_release和new_release中。 - 字符串
cook在old_release的版本中找到。 - 字符串
cook在版本中从new_release中找不到。我需要此条件，因为大多数代码未更新，并且将包含对cook的已停产引用。 - 我的存储库很大，检查一个分支需要很长时间。我想要一个避免这种情况的解决方案。
我目前的解决方案看起来像；
git checkout old_release
grep cook . -R -l -I > old_release_cooks.txt
sort -o old_release_cooks_sorted.txt old_release_cooks.txt
git checkout new_release
grep cook . -R -l -I > new_release_cooks.txt
sort -o new_release_cooks_sorted.txt new_release_cooks.txt
vim -d old_release_cooks_sorted.txt new_release_cooks_sorted.txt
除了第4点以外，这还满足我的所有要求。它要求我至少进行一次结帐。严格来说，我想它不会创建不同的文件列表，但是差异足够接近。
有没有办法在不签出的情况下获取此列表？
So, say I have two branches old_release (containing an old release) and new_release (containing a new release). In the old release there was a tool called cook, I think it's still in the new release, but it's been renamed. I need to work out what it's been renamed to.
I would like a list of files which must satisfy the criteria;
- File exists in both
old_releaseandnew_release. - String
cookis found in the version fromold_release. - String
cookis not found in the version fromnew_release. I need this condition because most of the code is not updated and will contain defunct references tocook. - My repository is big, and checking out a branch takes a long time. I'd like a solution that avoids that.
My current solution looks like;
git checkout old_release
grep cook . -R -l -I > old_release_cooks.txt
sort -o old_release_cooks_sorted.txt old_release_cooks.txt
git checkout new_release
grep cook . -R -l -I > new_release_cooks.txt
sort -o new_release_cooks_sorted.txt new_release_cooks.txt
vim -d old_release_cooks_sorted.txt new_release_cooks_sorted.txt
This meets all my requirements apart from point 4. It requires I do at least one checkout. Strictly I guess it doesn't create a list of files that differ, but the diff is close enough.
Is there a way to get this list without checking out?
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使用
git grep中的bash中的无用单线检查，并使用comm：word ='cook'rel1 ='old_release'rel2 ='new_release 评估其结果'; COMM -1 -1 -2＆lt;（git grep -l'\ b'“ $ word”'\ b'“ $ rel1” | sed's/^[^：]*：//'| sort;）＆lt;（ $ rel2” | sed's/^[^：]*：//'|git grep -l'\ b'“ $ word”'\ b' “ 条件1。满足（两者中必须存在文件）
edit2：简化的表达式
Checkout-free one-liner in bash that uses
git grepand evaluates its results usingcomm:word='cook' rel1='old_release' rel2='new_release'; comm -1 -2 <(git grep -l '\b'"$word"'\b' "$rel1" | sed 's/^[^:]*://' | sort;) <(git grep -L '\b'"$word"'\b' "$rel2" | sed 's/^[^:]*://' | sort;)EDIT: adjustments so that also your condition 1. is satisfied (file must exist in both)
EDIT2: simplified expression