如何在CMD批处理脚本中调用一个标签，当你使用超过9参数呢？ [英] How do you utilize more than 9 arguments when calling a label in a CMD batch-script?

问题描述

我想知道如何调用一个标签时批处理脚本中调用超过9说法。例如，下面的节目，我有12个参数分配以及试图呼应他们。

I would like to know how to call more than 9 argument within a batch script when calling a label. For example, the following shows that I have 12 arguments assigned along with attempting to echo all of them.

CALL:LABEL "one" "two" "three" "four" "five" "six" "seven" "eight" "nine" "ten" "eleven" "twelve"
PAUSE
GOTO:EOF


:LABEL
echo %1
echo %2
echo %3
echo %4
echo %5
echo %6
echo %7
echo %8
echo %9
echo %10
echo %11
echo %12

为10％，11％和％12的输出最终被one0 one1 one2。我已经使用大括号，括号，报价单，围绕数字单引号没有任何运气尝试。

The output for %10 %11 and %12 ends up being one0 one1 one2. I've tried using curly brackets, brackets, quotations, single quotes around the numbers without any luck.

推荐答案

有是 移该命令。

There is the shift command for that.

您可以使用一个循环，转移前的变量存储，或者做快速的是这样的：

You can either use a loop, store the variables before shifting, or do it quick like this:

CALL:LABEL "one" "two" "three" "four" "five" "six" "seven" "eight" "nine" "ten" "eleven" "twelve"
PAUSE
GOTO:EOF

:LABEL
echo %1
echo %2
echo %3
echo %4
echo %5
echo %6
echo %7
echo %8
echo %9

shift
shift 
shift 

echo %7
echo %8
echo %9

您可以在情况下，你有很多参数的循环更换换档命令。对于循环的执行移九次，使％1 将成为第十个参数。

You can replace the shift commands with a loop in case you have many arguments. The following for loop executes shift nine times, so that %1 will be the tenth argument.

@for /L %%i in (0,1,8) do shift

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