git add 并在一个命令中提交单个跟踪文件
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git add and commit single tracked file in one command
提问by BrodieG
I'm looking for the equivalent of git commit -am "blah blah"but for just a single file. If I try:
我正在寻找等效于git commit -am "blah blah"但仅用于单个文件的文件。如果我尝试：
git commit my.file -am "blah blah"
I get:
我得到：
fatal: Paths with -a does not make sense.
I looked around but I could only find solutions that suggest using aliases (e.g. this one), but even those don't seem like they could be modified because I need to pass an argument. Do I have to resort to calling git through a shell?
我环顾四周，但我只能找到建议使用别名的解决方案（例如这个），但即使是那些似乎也无法修改的解决方案，因为我需要传递一个参数。我必须求助于通过 shell 调用 git吗？
It seems like there should be a simpler option for something that I imagine would be extremely common. Right now I'm stuck with:
对于我认为非常普遍的事情，似乎应该有一个更简单的选择。现在我坚持：
git add my.file
git commit -m "blah blah"
回答by John Zwinck
Just omit the -awhich means --allwhich is not what you want. Do this:
只需省略-awhich 意味着--all这不是您想要的。做这个：
git commit my.file -m "blah blah"
That will commit only my.file, even if other files are staged (by git add).
my.file即使其他文件被暂存（by git add），那也只会提交。
回答by Sumit Verma
I tried the following command:
我尝试了以下命令：
git commit -m 'your comment' path/to/your/file.txt
I hope, this will help.
我希望这个能帮上忙。
