[英]How do I store the results of a shell command in a makefile?
Locally I'm usin GNU Make 3.8 on MacOS (i386-apple-darwin11.3.0) 我在本地使用MacOS上的GNU Make 3.8(i386-apple-darwin11.3.0)
I'm trying to generate a bunch of test files based on the files in a directory. 我正在尝试根据目录中的文件生成一堆测试文件。
Right now I'm able to grab the basename of the files, but I cant figure out how to store them in a variable so I can generate another file with the basename 现在,我可以获取文件的基本名称,但是我无法弄清楚如何将它们存储在变量中,因此可以生成另一个具有基本名称的文件
TEST_FILES = lib/*.js
build-test:
@mkdir -p tests
-@for file in $(TEST_FILES); do \
echo $$file; \
MYNAMES = $(basename $$file .js) \
echo $MYNAMES; \
done
I want to then create a bunch of files called $MYNAME.log using output from a STDOUT stream. 然后,我想使用STDOUT流的输出创建一堆名为$ MYNAME.log的文件。
I'm stuck on getting the names into the variable, the solutions I've found so far are throwing errors for me when I try to implement them 我一直坚持将名称放入变量中, 到目前为止 ,当我尝试实现它们时,我发现的解决方案正在为我抛出错误
This line here 这条线在这里
echo $MYNAMES; \
will substitute in the value of the make variable M
, as if you wrote $(M)YNAMES. 将替换make变量M
的值,就像您写了$(M)YNAMES一样。 You want 你要
echo $$MYNAMES; \
instead. 代替。
This line 这条线
MYNAMES = $(basename $$file .js) \
cannot work - you are using a make command basename
which is executed only once before the subshell is invoked to execute the rule, whereas file
is a shell variable which changes each iteration of the loop. 不能正常工作-您正在使用一个make命令的basename
,该basename
只能在调用子shell执行规则之前执行一次,而file
是一个shell变量,它会更改循环的每次迭代。 You probably meant $$(basename $$file)
instead, which will be passed to the shell as "$(basename $file)". 您可能是用$$(basename $$file)
代替的,它将作为“ $(basename $ file)”传递到外壳。 The shell will insert the results of running the basename
command. Shell将插入运行basename
命令的结果。
Make sure that you do want to store them in a shell variable, and not as a make variable. 确保确实要将它们存储在shell变量中,而不是作为make变量。 It's essential to understand the difference. 了解差异至关重要。
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