bash-通脚本作为另一个脚本的参数 [英] bash- pass script as argument of another script

问题描述

我不能于是找到了类似的问题。

I can't find a similar question on SO.

如何能够正确地传递一个bash脚本作为参数传递给另一个bash脚本。

How can I properly pass a bash script as an argument to another bash script.

例如，假设我有两个脚本，可以每接受多个参数，我想通过一个脚本，因为其他的争论。是这样的：

For example, let's say I have two scripts that can each accept a number of parameters, I want to pass one script as the argument of the other. Something like:

./script1 (./script2 file1 file2) file3

在上面的例子中，SCRIPT2合并文件1和file2在一起，回声一个新的文件，然而这是无关的问题。我只是想知道我怎么可以通过 SCRIPT2 作为一个参数，即正确的语法。

In the above example, script2 merges file1 and file2 together, and echos a new file, however that is irrelevant to the question. I just want to know how I can pass script2 as a parameter, i.e. the proper syntax.

如果这是不可能的，任何暗示，我怎么可以规避这个问题将是合适的。

If this is not possible, any hint as to how I may circumvent the issue would be appropriate.

推荐答案

如果你想传递的 SCRIPT2 作为参数传递给 SCRIPT1 执行它的最后一个里面，只要把下面的code里面的 SCRIPT1 并调用 SCRIPT1 是这样的：

If you want to pass script2 as an argument to script1 to execute it inside the last one, just put the following code inside script1 and call script1 like this:

./script1 "./script2 file1 file2" file3  # file4 file5

在

code SCRIPT1

$1 # here you're executing ./script2 file1 file2
shift
another_command "$@" # do anything else with the rest of params (file3)

---

或者，如果你知道PARAMS的要数 SCRIPT2 ，它是固定的，你也可以做到这一点如下：

---

Or if you know the number of params to script2 and it is fixed, you can also do it as follows:

./script1 ./script2 file1 file2 file3  # file4 file5

在

code SCRIPT1

"$1" "$2" "$3"
shift 3
another_command "$@" # do anything else with the rest of params (file3)

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