为每个GNU Makefile使用Basename

发布于 2025-01-19 21:23:09 字数 1177 浏览 3 评论 0原文

我想知道如何在每个内部使用basename？我认为Basename不起作用的原因可能是每个“ $ file”实际上是通往文件的途径，而不仅仅是文件名。

DOCKER_TAG=foobar
DOCKER_RUN=docker run -v `pwd`:/root -it --rm -w /root ${DOCKER_TAG}
TEST_DIR=test_cases

dockerbuild: App/App.csproj
    docker build . -t ${DOCKER_TAG}

%.run: %.txt dockerbuild 
    ${DOCKER_RUN} $<

%.out: %.txt dockerbuild
    ${DOCKER_RUN} $< > $@

%.test: %.out
    python3 evaluate.py $<

.PHONY: regression
regression: $(TEST_DIR)/*.txt
    for file in $^ ; do \
        echo $(basename $${file}).test ; \
    done

返回（我期望.txt已替换为.test）

test_cases/00_example.txt.test
test_cases/01_simplest_possible.txt.test
test_cases/02_two_users.txt.test
test_cases/03_five_users.txt.test
test_cases/04_one_interferer.txt.test
test_cases/05_equatorial_plane.txt.test
test_cases/06_partially_fullfillable.txt.test
test_cases/07_eighteen_planes.txt.test
test_cases/08_eighteen_planes_northern.txt.test
test_cases/09_ten_thousand_users.txt.test
test_cases/10_ten_thousand_users_geo_belt.txt.test
test_cases/11_one_hundred_thousand_users.txt.test
test_cases/test.txt.test

原文

I am wondering how to use basename inside for each? I think maybe the reason basename does not work is that each "$file" is actually a path to a file, not simply a filename.

DOCKER_TAG=foobar
DOCKER_RUN=docker run -v `pwd`:/root -it --rm -w /root ${DOCKER_TAG}
TEST_DIR=test_cases

dockerbuild: App/App.csproj
    docker build . -t ${DOCKER_TAG}

%.run: %.txt dockerbuild 
    ${DOCKER_RUN} lt;

%.out: %.txt dockerbuild
    ${DOCKER_RUN} lt; > $@

%.test: %.out
    python3 evaluate.py lt;

.PHONY: regression
regression: $(TEST_DIR)/*.txt
    for file in $^ ; do \
        echo $(basename ${file}).test ; \
    done

Returns (I expected .txt to have been replaced with .test)

test_cases/00_example.txt.test
test_cases/01_simplest_possible.txt.test
test_cases/02_two_users.txt.test
test_cases/03_five_users.txt.test
test_cases/04_one_interferer.txt.test
test_cases/05_equatorial_plane.txt.test
test_cases/06_partially_fullfillable.txt.test
test_cases/07_eighteen_planes.txt.test
test_cases/08_eighteen_planes_northern.txt.test
test_cases/09_ten_thousand_users.txt.test
test_cases/10_ten_thousand_users_geo_belt.txt.test
test_cases/11_one_hundred_thousand_users.txt.test
test_cases/test.txt.test

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醉生梦死 2025-01-26 21:23:09

$(basename) 函数在 shell 运行循环之前由 make 求值。因此，您有效地运行了

for file in /path/to/moo.txt /path/to/bar.txt /path/to/baz.txt; do
    echo ${file}.test ;
done

shell 的 basename 命令也会删除路径，但您可能想用类似的方法来修复此问题

.PHONY: regression
regression: $(TEST_DIR)/*.txt
    printf '%s\n' $^ | sed 's/\.txt$/.test/'

，当然，更好的解决方案是使用 make 函数反而。

.PHONY: regression
regression: $(TEST_DIR)/*.txt
    printf '%s\n' $(patsubst %,%.test,$(basename $^))

如果您想做的不仅仅是打印文件名，更惯用的解决方案是创建一个模式规则并在每个输入文件上运行它。

.PHONY: regression
regression: $(patsubst %.txt,%.test,$(wildcard $(TEST_DIR)/%.txt))

$(TEST_DIR)/%.test: $(TEST_DIR)/%.txt
    printf '%s\n' $@ # or whatever you really wanted to do

The $(basename) function is evaluated by make before the shell runs the loop. So you are effectively running

for file in /path/to/moo.txt /path/to/bar.txt /path/to/baz.txt; do
    echo ${file}.test ;
done

The shell's basename command also removes the path, but you might want to fix this with something like

.PHONY: regression
regression: $(TEST_DIR)/*.txt
    printf '%s\n' $^ | sed 's/\.txt$/.test/'

though of course, a better solution is to use make functions instead.

.PHONY: regression
regression: $(TEST_DIR)/*.txt
    printf '%s\n' $(patsubst %,%.test,$(basename $^))

If you want to do something more than just print the file names, a more idiomatic solution is to create a pattern rule and run it on each of the input files.

.PHONY: regression
regression: $(patsubst %.txt,%.test,$(wildcard $(TEST_DIR)/%.txt))

$(TEST_DIR)/%.test: $(TEST_DIR)/%.txt
    printf '%s\n' $@ # or whatever you really wanted to do

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