英文:
Passing Makefile variable to docker-compose is resulting in a "unknown shorthand flag" error
问题
我有一个Makefile，我用它来帮助我们的程序员轻松设置环境。目前，我正在编写一个Makefile来将IMAGE标签传递给docker-compose命令，但我一直收到一个unknown shorthand flag错误，我无法弄清楚原因。
我的Makefile如下：
.PHONY: allARCH = $(shell uname -m)start:if [ "$(ARCH)" = "x86_64" ]; then \IMAGE_ARCH="amd64"; \elif [ "$(ARCH)" = "aarch64" ]; then \IMAGE_ARCH="arm64"; \else \echo "Unknown architecture: $(ARCH)"; \fidocker-compose up -d -e IMAGE=$IMAGE_ARCH
错误信息是：
unknown shorthand flag: 'e' in -e
英文:
I have a Makefile which I use to help our programmers get easily setup. At present I am writing one to pass an IMAGE tag to the docker-compose command, but I keep getting a unknown shorthand flag and I cant figure out why.
My Makefile is as follows;
.PHONY: allARCH = $(shell uname -m)start:if [ "$(ARCH)" = "x86_64" ]; then \IMAGE_ARCH="amd64"; \elif [ "$(ARCH)" = "aarch64" ]; then \IMAGE_ARCH="arm64"; \else \echo "Unknown architecture: $(ARCH)"; \fidocker-compose up -d -e IMAGE=$IMAGE_ARCH
The error is;
unknown shorthand flag: 'e' in -e
答案1
得分: 1
docker-compose没有-e选项。
你可以这样运行：
.PHONY: allARCH := $(shell uname -m)start:if [ "$(ARCH)" = "x86_64" ]; then \IMAGE_ARCH="amd64"; \elif [ "$(ARCH)" = "aarch64" ]; then \IMAGE_ARCH="arm64"; \else \echo "Unknown architecture: $(ARCH)"; \fi; \IMAGE=$$IMAGE_ARCH docker-compose up -d
docker-compose命令需要放在与[if]相同的块中，以便使用$IMAGE_ARCH。
英文:
docker-compose has no -e option.
You can run this way :
.PHONY: allARCH := $(shell uname -m)start:if [ "$(ARCH)" = "x86_64" ]; then \IMAGE_ARCH="amd64"; \elif [ "$(ARCH)" = "aarch64" ]; then \IMAGE_ARCH="arm64"; \else \echo "Unknown architecture: $(ARCH)"; \fi; \IMAGE=$$IMAGE_ARCH docker-compose up -d
docker-compose command needs to be put in the same block as [if] in order to use $IMAGE_ARCH.
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