shell - AWK 用于逐行检查不同文件的第一列并查看差异

Question

我有 2 个这样的文件:

文件1:

fa65c95d0  ./checklist.chk
316856ab2  ./.vscode/settings.json
6075df9cc  ./sha256sum.txt
36d59e45b  ./md5sum.txt

文件2:

fa65c95d0  ./checklist.chk
222222222  ./.vscode/settings.json
644ddd555  ./sha256sum.txt
36d59e45b  ./md5sum.txt

我想逐行比较这些文件的第一列，看看 File2 有什么不同。

期望的输出

./.vscode/settings.json is modified
./sha256sum.txt is modified

我在网上搜索并尝试使用 comm，但是这个脚本并不完全符合我的要求

comm -3 sum1.txt sum2.txt | awk '$1 {print $2}'

为此输出

comm: file 1 is not in sorted order
comm: file 2 is not in sorted order
./.vscode/settings.json
./.vscode/settings.json
./sha256sum.txt
./md5sum.txt
./sha256sum.txt
./md5sum.txt

Answer 1

使用diff(1):

diff file1 file2

2,3c2,3
< 316856ab2  ./.vscode/settings.json
< 6075df9cc  ./sha256sum.txt
---
> 222222222  ./.vscode/settings.json
> 644ddd555  ./sha256sum.txt

或统一差异:

diff -u file1 file2

--- file1   2019-12-02 18:44:46.525142623 +0100
+++ file2   2019-12-02 18:44:57.253178157 +0100
@@ -1,4 +1,4 @@
 fa65c95d0  ./checklist.chk
-316856ab2  ./.vscode/settings.json
-6075df9cc  ./sha256sum.txt
+222222222  ./.vscode/settings.json
+644ddd555  ./sha256sum.txt
 36d59e45b  ./md5sum.txt

或者vimdiff(1):

vimdiff file1 file2