是否有一种方法来指示批处理文件中的最后n个参数? [英] Is there a way to indicate the last n parameters in a batch file?
问题描述
在下面的示例中,我想从父批处理文件调用子批处理文件,并将所有剩余的参数传递给子进程。
In the following example, I want to call a child batch file from a parent batch file and pass all of the remaining parameters to the child.
C:\> parent.cmd child1 foo bar
C:\> parent.cmd child2 baz zoop
C:\> parent.cmd child3 a b c d e f g h i j k l m n o p q r s t u v w x y z
在parent.cmd内,我需要删除%1参数,并且只将剩余的参数传递给子脚本。
Inside parent.cmd, I need to strip %1 off the list of parameters and only pass the remaining parameters to the child script.
set CMD=%1
%CMD% <WHAT DO I PUT HERE>
我使用SHIFT和%*进行了调查,但是不行。虽然SHIFT会将位置参数向下移动1,但%*仍指原始参数。
I've investigated using SHIFT with %*, but that doesn't work. While SHIFT will move the positional parameters down by 1, %* still refers to the original parameters.
任何人都有任何想法?我应该放弃并安装Linux吗?
Anyone have any ideas? Should I just give up and install Linux?
推荐答案
%*
总是扩大到所有的原始参数,可悲的是。但是你可以使用下面的代码片段来构建一个包含第一个参数的变量:
%*
will always expand to all original parameters, sadly. But you can use the following snippet of code to build a variable containing all but the first parameter:
rem throw the first parameter away
shift
set params=%1
:loop
shift
if [%1]==[] goto afterloop
set params=%params% %1
goto loop
:afterloop
,虽然...我不经常写这样的事情:)
I think it can be done shorter, though ... I don't write these sort of things very often :)
但应该工作。
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