如何使下面的makefile条件工作

Question

FOLDER_NAME = $*

%.clean:
ifeq ($(FOLDER_NAME),cdr)
    @ echo " condition passed $(FOLDER_NAME)"
else
    @ echo " condition failed $(FOLDER_NAME)"
endif

在此运行make cdr.clean时，我得到以下输出。 条件失败的cdr

无法弄清楚可能出现的问题。

Answer 1

读取makefile时会计算条件ifeq。那时$*评估为空字符串，因此ifeq看到空FOLDER_NAME。 $*仅在以后在运行时评估规则时才有意义。

为了实现您的目标，您可以为不同的情况定义不同的规则（* clean2如下）或使用shell命令来评估条件（* clean3如下）。

FOLDER_NAME = $*

cdr.clean2:
        @ echo " condition passed $(FOLDER_NAME)"

%.clean2:
        @ echo " condition failed $(FOLDER_NAME)"

%.clean3:
        @ if [ $(FOLDER_NAME) == cdr ]; \
            then echo " condition passed $(FOLDER_NAME)" ; \
            else echo " condition failed $(FOLDER_NAME)"; \
          fi