Início / coding / Perguntas / 79560439
Accepted
vbulash
Asked: 2025-04-08 01:02:15 +0800 CST

Condição do Makefile `ifeq` - o que está errado

  • 772

Meu makefile é:

LOGGER=Mongo
# Mongo logger
LOGGER_MONGO_BINARY=loggerMongoApp
LOGGER_MONGO_FOLDER=logger-service-mongo
# Redis logger
LOGGER_REDIS_BINARY=loggerRedisApp
LOGGER_REDIS_FOLDER=logger-service-redis

build_logger:
    ifeq ("${LOGGER}", "Mongo")
    LOGGER_BINARY=${LOGGER_MONGO_BINARY}
    LOGGER_FOLDER=${LOGGER_MONGO_FOLDER}
    else ifeq ("${LOGGER}", "Redis")
    LOGGER_BINARY=${LOGGER_REDIS_BINARY}
    LOGGER_FOLDER=${LOGGER_REDIS_FOLDER}
    endif

    @echo Building logger binary (${LOGGER}-based)...
    chdir ..\${LOGGER_FOLDER} \
        && set GOOS=linux \
        && set GOARCH=amd64 \
        && set CGO_ENABLED=0 \
        && go build -o ${LOGGER_BINARY} ./cmd/api
    @echo Done!

No makefile quero obter lógica condicional.

make build_loggertermina com:

ifeq ("Mongo", "Mongo")
process_begin: CreateProcess(NULL, ifeq (Mongo, Mongo), ...) failed.
make (e=2): ═х єфрхЄё  эрщЄш єърчрээ√щ Їрщы.
make: *** [Makefile:43: build_logger] Error 2

A mensagem de texto dentro é "não foi possível encontrar o arquivo especificado". O que há de errado?

makefile

1 respostas

  1. Best Answer

    Recomendo que você escreva assim:

    LOGGER = Mongo

# Mongo logger
Mongo_LOGGER_BINARY = loggerMongoApp
Mongo_LOGGER_FOLDER = logger-service-mongo

# Redis logger
Redis_LOGGER_BINARY = loggerRedisApp
Redis_LOGGER_FOLDER = logger-service-redis

LOGGER_BINARY = $($(LOGGER)_LOGGER_BINARY)
LOGGER_FOLDER = $($(LOGGER)_LOGGER_FOLDER)

build_logger:
        @echo Building logger binary (${LOGGER}-based)...
        chdir ..\${LOGGER_FOLDER} \
            && set GOOS=linux \
            && set GOARCH=amd64 \
            && set CGO_ENABLED=0 \
            && go build -o ${LOGGER_BINARY} ./cmd/api
        @echo Done!
    • 0

relate perguntas