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相同的 make 命令在 macOS 中失败，但在 Linux 中成功

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相同的 make 命令在 macOS 中失败，但在 Linux 中成功

我在下面使用Makefile，make publish在 macOS 上失败，而在 Linux（alpine、Ubuntu、Debian）上成功，并出现以下错误：

$ make publish
"docker images" requires at most 1 argument.
See 'docker images --help'.

Usage:  docker images [OPTIONS] [REPOSITORY[:TAG]]

List images
"docker inspect" requires at least 1 argument.
See 'docker inspect --help'.

Usage:  docker inspect [OPTIONS] NAME|ID [NAME|ID...]

Return low-level information on Docker objects
=> Publishing release image to some_image:tag ...
=> Publish complete

make publishsome_image:tag使用.push 推送之前构建的 Docker 镜像make image_build。

生成文件如下：

ORG_NAME ?= some_name
REPO_NAME ?= some_name
 
CHECK := @bash -c '\
  if [[ $(INSPECT) -ne 0 ]]; \
  then exit $(INSPECT); fi' VALUE

DOCKER_REGISTRY ?= [some_docker_registry]

.PHONY: build image_build tag login logout publish

build:
    ${INFO} "Building artifacts..." 
    ${INFO} "Building artifacts complete"

image_build:
    ${INFO} "Building images..."
    @ docker build -t $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME) .
    ${INFO} "Building complete..."

tag:
    ${INFO} "Tagging release image with tags $(TAG_ARGS)..."
    @ $(foreach tag,$(TAG_ARGS), docker tag $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME):latest $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME):$(tag);)
    ${INFO} "Tagging complete"  
 
login:
    ${INFO} "Logging in to Docker registry $$DOCKER_REGISTRY..."
    @ docker login -u $$DOCKER_USER -p $$DOCKER_PASSWORD $(DOCKER_REGISTRY)
    ${INFO} "Logged in to Docker registry $$DOCKER_REGISTRY"
 
logout:
    ${INFO} "Logging out of Docker registry $$DOCKER_REGISTRY..."
    @ docker logout $(DOCKER_REGISTRY)
    ${INFO} "Logged out of Docker registry $$DOCKER_REGISTRY"
 
publish:
    ${INFO} "Publishing release image to $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME)..."    
    @ $(foreach tag,$(shell echo $(REPO_EXPR)), docker push $(tag);)
    ${INFO} "Publish complete"

IMAGE_ID := $$(docker images -q $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME):latest)

# Repository Filter
ifeq ($(DOCKER_REGISTRY), docker.io)
    REPO_FILTER := $(ORG_NAME)/$(REPO_NAME)[^[:space:]|\$$]*
else
    REPO_FILTER := $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME)[^[:space:]|\$$]*
endif

# Introspect repository tags
REPO_EXPR := $$(docker inspect -f '{{range .RepoTags}}{{.}} {{end}}' $(IMAGE_ID) | grep -oh "$(REPO_FILTER)" | xargs)

# Cosmetics
YELLOW := "\e[1;33m"
NC := "\e[0m"
 
# Shell Functions
INFO := @bash -c '\
  printf $(YELLOW); \
  echo "=> $$1"; \
  printf $(NC)' SOME_VALUE

# Extract tag arguments
ifeq (tag,$(firstword $(MAKECMDGOALS)))
  TAG_ARGS := $(wordlist 2,$(words $(MAKECMDGOALS)),$(MAKECMDGOALS))
  ifeq ($(TAG_ARGS),)
    $(error You must specify a tag)
  endif
  $(eval $(TAG_ARGS):;@:)
endif

想知道如何调试此类问题以及可能的原因是什么？

latest尝试从 line 中删除IMAGE_ID := $$(docker images -q $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME):latest)，但没有任何反应：

$ make publish
=> Publishing release image to some_image:tag
=> Publish complete

makemacOS 版本：

$ make --version
GNU Make 3.81
Copyright (C) 2006  Free Software Foundation, Inc.
This is free software; see the source for copying conditions.
There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A
PARTICULAR PURPOSE.

This program built for i386-apple-darwin11.3.0

答案1

编辑

实际上，在再次查看 Makefile 后，我注意到“docker”命令抱怨缺少参数，这意味着它无法找到您之前构建的映像

publish:
    ${INFO} "Publishing release image to $(DOCKER_REGISTRY)/$(ORG_NAME)/$(REPO_NAME)..."
    ${INFO} "$(IMAGE_ID)"
    @ $(foreach tag,$(shell echo $(REPO_EXPR)), docker push $(tag);)

 $ make publish
"docker inspect" requires at least 1 argument.
See 'docker inspect --help'.

Usage:  docker inspect [OPTIONS] NAME|ID [NAME|ID...]

Return low-level information on Docker objects
=> Publishing release image to ****...
=>
=> Publish complete

如果图像存在，则发布步骤将正常工作

 $ make publish
=> Publishing release image to ***** ...
The push refers to repository [*****k]
90f9b3de67de: Pushed
latest: digest: sha256:af045c36b459f251b41b2db6b98320b7fdc1d6a2146b17a15344ec0e0a89b6f5 size: 527
=> Publish complete

所以你需要确保提前执行“image_build”步骤

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