dongyi7513 2017-07-23 02:54 采纳率: 100%
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从项目Makefile检测GOPATH

If GOPATH is not set, compilation for go programs is impossible. But many go projects are built using Makefiles, because go also miss capabilities to extract git revision, set version etc. So it should be possible to detect GOPATH from Makefile automatically.

Suppose I set my GOPATH manually one time for go get -d:

go get -d github.com/zyedidia/micro/cmd/micro

Now if I open another session, cd into github.com/zyedidia/micro/cmd/micro and do make build, the build fails:

...
cmd/micro/micro.go:20:2: cannot find package "layeh.com/gopher-luar" in any of:
    /usr/lib/go-1.7/src/layeh.com/gopher-luar (from $GOROOT)
    ($GOPATH not set)
Makefile:15: recipe for target 'build' failed
make: *** [build] Error 1

So, if GOPATH is not set, how can I set it from the Makefile and make sure that there is go environment at this point?

This doesn't work:

GOPATH ?= ../../../..

UPDATE: The following code works, but it doesn't detect that parent directory contains src, bin and pkg dirs.

export GOPATH ?= $(abspath $(dir ../../../../..))

export is needed to transform make variable into environment variable, ?= sets make variable only of it is not set, abspath and dir are described here:

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  • douxian7117 2017-07-23 07:09
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    Here is the solution.

    # detect GOPATH if not set
    ifndef $(GOPATH)
        $(info GOPATH is not set, autodetecting..)
        TESTPATH := $(dir $(abspath ../../..))
        DIRS := bin pkg src
        # create a ; separated line of tests and pass it to shell
        MISSING_DIRS := $(shell $(foreach entry,$(DIRS),test -d "$(TESTPATH)$(entry)" || echo "$(entry)";))
        ifeq ($(MISSING_DIRS),)
            $(info Found GOPATH: $(TESTPATH))
            export GOPATH := $(TESTPATH)
        else
            $(info ..missing dirs "$(MISSING_DIRS)" in "$(TESTDIR)")
            $(info GOPATH autodetection failed)
        endif
    endif
    

    What I've learned:

    • variables are defines in a separate block
    • tabs are not allowed in the block that defines variables
    • echo doesn't work in this block, need to use $(info)
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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