遍历所有的参数数组，不知道有多少争论，在shell脚本？ [英] Loop over the arguments array, without knowing how many arguments, in a shell script?

问题描述

我想很多参数传递给shell脚本，我不知道他们有多少论点将是，我要来处理它们。我做了以下code：

I want to pass many arguments to a shell script which I don't know how many arguments they are going to be and I want to handle them. I did the following code:

int=$1
src=$2

r=$3
string=$4

duration=$5

./start.sh $int $r $src "$string"
sleep $duration

shift; shift; shift; shift; shift

while [ $# -gt 2 ]
do
  r=$1
  string=$2
  duration=$3
  ./change.sh $int $r "$string"
  sleep $duration
  shift; shift; shift
done

这工作，但只对一次，现在我想这个脚本来运行所有的时间，我的意思是使用，而1 但不会以这种方式工作，因为参数列表就是在那个code结束时清空！

That works but only for one time, now I want this script to run all the time, I mean using while 1 but that won't work this way because the argument list is empty at the end of that code!

有没有办法做类似下面的shell脚本伪code：

Is there any way to do something like the following "pseudo code" in shell scripts:

for( i=0 ; i<arguments.count; i++ ){
   //do something with arguments[i]
}

或复制参数数组到另一个数组，这样我可以在以后使用它，我想要的方式。
我的意思是我可以复制 $ * 或 $ @ 或 ARG 到另一个阵列？

or copying the arguments array into another array so that I can use it later the way I want. I mean can I copy $* or $@ or arg into another array?

任何帮助是非常AP preciated。

Any help is highly appreciated.

推荐答案

参数的个数存储在可与<$ C $访问参数＃， C> $＃。

The number of arguments is stored in the parameter #, which can be accessed with $#.

在所有参数的简单循环可以写成如下：

A simple loop over all arguments can be written as follows:

for arg
do
    # do something with the argument "arg"
done

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