强制Makefile在构建目标之前执行脚本 [英] Force Makefile to execute script before building targets

问题描述

我正在使用Makefile.

I am using Makefiles.

但是，在执行任何目标之前，有一个我要执行的命令(zsh脚本). 我该怎么做?

However, there is a command (zsh script) I want executed before any targets is executed. How do I do this?

谢谢！

推荐答案

有几种技术可以在构建目标之前执行代码.您应该选择哪一个取决于您想做什么，以及为什么要这样做. (zsh脚本有什么作用?为什么必须执行它?)

There are several techniques to have code executed before targets are built. Which one you should choose depends a little on exactly what you want to do, and why you want to do it. (What does the zsh script do? Why do you have to execute it?)

您可以按照@John的建议进行操作；将zsh脚本放置为第一个依赖项.然后，应将zsh目标标记为.PHONY，除非它实际上生成了名为zsh的文件.

You can either do like @John suggests; placing the zsh script as the first dependency. You should then mark the zsh target as .PHONY unless it actually generates a file named zsh.

另一种解决方案(至少在GNU make中)是调用$(shell ...)函数作为变量赋值的一部分:

Another solution (in GNU make, at least) is to invoke the $(shell ...) function as part of a variable assignment:

ZSH_RESULT:=$(shell zsh myscript.zsh)

这将在解析makefile之后且在执行任何目标之前执行脚本.如果您递归调用makefile，它也会执行脚本.

This will execute the script as soon as the makefile is parsed, and before any targets are executed. It will also execute the script if you invoke the makefile recursively.

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