perforce - 是否有可能从一个变更列表中减去另一个变更列表的变更？

Question

假设我有两个变更列表 A 和 B(位于 A 之上)。
现在，我只想要 B 中所做的更改，而不是 A 中包含的更改，即 B-A
可以这样做吗？

Answer 1

是的。做:

p4 undo @A,A
p4 sync @B

这将创建一个新的待定更改列表，即“B 减去 A”。如果 A 和 B 接触了相同的文件，则同步到 @B 将需要解析将 B 的更改合并到文件中(没有 A 的更改)。合并的设置方式将使一切尽可能自动发生。

如果以后对同一文件进行了其他更改，您将需要同步并解决这些更改，就像您在任何时候所做的那样正在处理旧的变更列表。

理论上这在 P4V 中也是可行的(我认为这是“退出”命令)，但我对此有过不好的经历，因此不推荐它。从命令行非常简单。

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示例

C:\Perforce\test\undo>p4 annotate -c foo
//stream/main/undo/foo#3 - edit change 137 (text)
135: this
136: is    **CHANGE A**
135: a
137: test  **CHANGE B**
135: file

更改 A 为 @136，更改 B 为 @137。

C:\Perforce\test\undo>p4 undo @136,136
//stream/main/undo/foo#2 - opened for integrate
... undid //stream/main/undo/foo#2

C:\Perforce\test\undo>p4 sync @137
//stream/main/undo/foo#3 - is opened and not being changed
... //stream/main/undo/foo - must resolve #3 before submitting

C:\Perforce\test\undo>p4 resolve -am
c:\Perforce\test\undo\foo - merging //stream/main/undo/foo#3
Diff chunks: 1 yours + 1 theirs + 0 both + 0 conflicting
//compy386/undo/foo - merge from //stream/main/undo/foo

C:\Perforce\test\undo>p4 submit -d "undid A"
Submitting change 138.
Locking 1 files ...
integrate //stream/main/undo/foo#4
Change 138 submitted.

更改 136 现已被更改 138 撤消:

C:\Perforce\test\undo>p4 annotate -c foo
//stream/main/undo/foo#4 - integrate change 138 (text)
135: this
138: is
135: a
137: test  **CHANGE B**
135: file

C:\Perforce\test\undo>p4 filelog foo
//stream/main/undo/foo
... #4 change 138 integrate on 2020/02/08 by Samwise@compy386 (text) 'undid A'
... ... undid //stream/main/undo/foo#2
... #3 change 137 edit on 2020/02/08 by Samwise@compy386 (text) 'change B'
... #2 change 136 edit on 2020/02/08 by Samwise@compy386 (text) 'change A'
... ... undone by //stream/main/undo/foo#4
... #1 change 135 add on 2020/02/08 by Samwise@compy386 (text) 'test file'