Git diff:仅显示与模式不匹配的更改
是否可以让git diff假定以某种模式凝视的线条没有改变?
Is it possible to tell git diff to assume lines staring with some pattern as unchanged?
例如，考虑以下内容:
$ git diff -U0
diff --git a/file_a.txt b/file_a.txt
index 26ed843..4071ff8 100644
--- a/file_a.txt
+++ b/file_a.txt
@@ -24 +24 @@
- * unimportant foo
+ * unimportant bar
diff --git a/file_b.txt b/file_b.txt
index c6d051e..4b3cf22 100644
--- a/file_b.txt
+++ b/file_b.txt
@@ -24 +24 @@
- * unimportant foo
+ * unimportant bar
@@ -48,0 +49 @@
+ this is important
@@ -56,0 +58 @@
+ this is also important
以星号开头的行(正则表达式模式"^[[:space:]]*\*.*")并不重要，我想仅从git diff的输出中过滤出包含此类行更改的文件.在上面的示例中，输出仅报告file_b.txt更改.有可能吗?
Lines starting with an asterisk (regex pattern "^[[:space:]]*\*.*") are not important and I would like to filter files that contain changes in such lines only from the output of git diff. In the example above, the output should report file_b.txt changes only. Is it possible?
可以使用git diff -G标志并反转正则表达式以匹配行，即第一个非空格字符既不是*也不是空格.
It's possible with the git diff -G flag and making inverting the regexp to match lines that the first character that is not space is neither a * nor space.
-G '^[[:space:]]*[^[:space:]*]'
效率不高，因为会回溯，但不支持负超前'^(?!\s*\*)'，所有格修饰符'^\s*+[^*]'或原子组'^(?>\s*)[^*]'.
Not very efficient because will backtrack but seems negative lookahead '^(?!\s*\*)', possessive quantifier '^\s*+[^*]' or atomic groups '^(?>\s*)[^*]' are not supported.