英文:
How do I wildcard a rule with SECONDEXPANSION in GNU Make?
问题
# Docker Chain
src_files:=$(shell find src/ -name '*.py')
image_tags=$(shell find dkr/ -name '$(PROJECT).*.dockerfile' | xargs -n 1 basename -s '.dockerfile')
dkr/%.dockerfile: $(src_files)
docker build . -f $@ -t $*
.SECONDEXPANSION:
$(PROJECT).base: dkr/$$@.dockerfile
.SECONDEXPANSION:
$(PROJECT).%: $(PROJECT).base | dkr/$$*.dockerfile
publish-%:
$(MAKE) $*
./scripts/$@.sh $(RESOURCE_PREFIX) $* $(REGION)
publish:
$(foreach tag, $(image_tags), $(MAKE) publish-$(tag))
英文:
# Docker Chain
src_files:=$(shell find src/ -name '*.py')
image_tags=$(shell find dkr/ -name '$(PROJECT).*.dockerfile' | xargs -n 1 basename -s '.dockerfile')
dkr/%.dockerfile: $(src_files)
docker build . -f $@ -t $*
.SECONDEXPANSION:
$(PROJECT).base: dkr/$$@.dockerfile
.SECONDEXPANSION:
$(PROJECT).foo: $(PROJECT).base | dkr/$$@.dockerfile
.SECONDEXPANSION:
$(PROJECT).bar: $(PROJECT).base | dkr/$$@.dockerfile
publish-%:
$(MAKE) $*
./scripts/$@.sh $(RESOURCE_PREFIX) $* $(REGION)
publish:
$(foreach tag, $(image_tags), $(MAKE) publish-$(tag))
I would like to write the above as:
# Docker Chain
src_files:=$(shell find src/ -name '*.py')
image_tags=$(shell find dkr/ -name '$(PROJECT).*.dockerfile' | xargs -n 1 basename -s '.dockerfile')
dkr/%.dockerfile: $(src_files)
docker build . -f $@ -t $*
.SECONDEXPANSION:
$(PROJECT).base: dkr/$$@.dockerfile
.SECONDEXPANSION:
$(PROJECT).%: $(PROJECT).base | dkr/$$@.dockerfile
publish-%:
$(MAKE) $*
./scripts/$@.sh $(RESOURCE_PREFIX) $* $(REGION)
publish:
$(foreach tag, $(image_tags), $(MAKE) publish-$(tag))
However, the wildcarding in the rule (via '%') does not work. I suppose I have misunderstood rule substitution, and am guessing it is due to the SECONEXPANSION.
But, my make intuition is telling me this should work, and it should be good.
How would I perform substitution in this scenario?
答案1
得分: 0
代码上的问题出在上面，而不是“其他地方”。此外，你对.SECONDEXPANSION的应用是正确的；问题不在那里。在这里：
在Makefile中，你可以有一个依赖规范，而不需要为其提供任何配方，就像你的第一个例子中一样：
$(PROJECT).foo: $(PROJECT).base | dkr/$$@.dockerfile
但是，如果你尝试使用模式（%）来创建这样的依赖关系（没有配方），你就无法实现（Make使用这样的东西来表示其他东西）。要创建“模式规则”，你必须提供一个配方。如果你实际上没有一个配方，那就提供虚拟的配方，即在依赖关系行之后加上分号。
所以
$(PROJECT).%: $(PROJECT).base | dkr/$$@.dockerfile ;
这将解决你的问题。
顺便说一下，我不明白为什么你需要那些“顺序前提条件”（在|之后）。它们用于防止目标仅因为先决条件更新而重新构建。由于这些目标没有配方，因此没有必要设置顺序前提条件。所以只需执行
$(PROJECT).%: $(PROJECT).base dkr/$$@.dockerfile ;
英文:
The problem is in the code above, not "elsewhere". Also, your application of .SECONDEXPANSION is correct; the problem is not there. It is here:
You can have a dependency specification in a makefile, without any recipe to it, like in your first example:
$(PROJECT).foo: $(PROJECT).base | dkr/$$@.dockerfile
But, if you try to have such a dependency (without a recipe), with a pattern (%), you can't (Make uses such a thing for something else). To have a "pattern rule", you must supply a recipe. If you don't really have a recipe, then supply the dummy recipe, which is a semicolon after the dependency line.
So
$(PROJECT).%: $(PROJECT).base | dkr/$$@.dockerfile ;
and that will solve your problem.
By the way, I don't see why you need those "order prerequisites" (after the |). They are used to prevent the target from rebuilding solely because the prerequisite is newer. Since these targets do not have a recipe, there is no point in having order prerequisites. So just do
$(PROJECT).%: $(PROJECT).base dkr/$$@.dockerfile ;
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