Linux 如何在 Makefile 中将 dir 添加到 $PATH 中？
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How I could add dir to $PATH in Makefile?
提问by Szymon Lipiński
I want to write a Makefile which would run tests. Test are in a directory './tests' and executable files to be tested are in the directory './bin'.
我想编写一个可以运行测试的 Makefile。测试在目录“./tests”中，要测试的可执行文件在目录“./bin”中。
When I run the tests, they don't see the exec files, as the directory ./bin is not in the $PATH.
当我运行测试时，他们看不到 exec 文件，因为目录 ./bin 不在 $PATH 中。
When I do something like this:
当我做这样的事情时：
EXPORT PATH=bin:$PATH
make test
everything works. However I need to change the $PATH in the Makefile.
一切正常。但是我需要更改 Makefile 中的 $PATH。
Simple Makefile content:
简单的 Makefile 内容：
test all:
PATH=bin:${PATH}
@echo $(PATH)
x
It prints the path correctly, however it doesn't find the file x.
它正确打印路径，但是它没有找到文件 x。
When I do this manually:
当我手动执行此操作时：
$ export PATH=bin:$PATH
$ x
everything is OK then.
一切正常。
How could I change the $PATH in the Makefile?
如何更改 Makefile 中的 $PATH？
采纳答案by Eldar Abusalimov
Did you try exportdirectiveof Make itself (assuming that you use GNU Make)?
您是否尝试过 Make 本身的export指令（假设您使用 GNU Make）？
export PATH := bin:$(PATH)
test all:
x
Also, there is a bug in you example:
此外，您的示例中有一个错误：
test all:
PATH=bin:${PATH}
@echo $(PATH)
x
First, the value being echoed is an expansion of PATHvariable performed by Make, not the shell. If it prints the expected value then, I guess, you've set PATHvariable somewhere earlier in your Makefile, or in a shell that invoked Make. To prevent such behavior you should escape dollars:
首先，被echoed的值是PATH由 Make 执行的变量的扩展，而不是 shell。如果它打印出预期值，那么我猜你已经PATH在 Makefile 的某个地方或在调用 Make 的 shell 中的某个地方设置了变量。为了防止这种行为，你应该逃避美元：
test all:
PATH=bin:$$PATH
@echo $$PATH
x
Second, in any case this won't work because Make executes each line of the recipe in a separate shell. This can be changed by writing the recipe in a single line:
其次，无论如何这都行不通，因为 Make 在单独的 shell 中执行配方的每一行。这可以通过在一行中编写配方来更改：
test all:
export PATH=bin:$$PATH; echo $$PATH; x
回答by karnhick
To set the PATHvariable, within the Makefile only, use something like:
要设置PATH变量，仅在 Makefile 中，使用类似的内容：
PATH := $(PATH):/my/dir
test:
@echo my new PATH = $(PATH)
回答by Richard Pennington
What I usually do is supply the path to the executable explicitly:
我通常做的是明确提供可执行文件的路径：
EXE=./bin/
...
test all:
$(EXE)x
I also use this technique to run non-native binaries under an emulator like QEMU if I'm cross compiling:
如果我进行交叉编译，我还使用这种技术在 QEMU 等模拟器下运行非本地二进制文件：
EXE = qemu-mips ./bin/
If make is using the sh shell, this should work:
如果 make 正在使用 sh shell，这应该可以工作：
test all:
PATH=bin:$PATH x
回答by underspecified
Path changes appear to be persistent if you set the SHELL variable in your makefile first:
如果您首先在 makefile 中设置 SHELL 变量，则路径更改似乎是持久的：
SHELL := /bin/bash
PATH := bin:$(PATH)
test all:
x
I don't know if this is desired behavior or not.
我不知道这是否是理想的行为。
回答by kenorb
By design makeparser executes lines in a separate shell invocations, that's why changing variable (e.g. PATH) in one line, the change may not be applied for the next lines (see this post).
根据设计，make解析器在单独的 shell 调用中执行行，这就是为什么PATH在一行中更改变量（例如），更改可能不会应用于下一行（请参阅此帖子）。
One way to workaround this problem, is to convert multiple commands into a single line (separated by ;), or use One Shellspecial target (.ONESHELL, as of GNU Make 3.82).
解决此问题的一种方法是将多个命令转换为一行（以 分隔;），或使用One Shell特殊目标（.ONESHELL，从 GNU Make 3.82 开始）。
Alternatively you can provide PATHvariable at the time when shell is invoked. For example:
或者，您可以PATH在调用 shell 时提供变量。例如：
PATH := $(PATH):$(PWD)/bin:/my/other/path
SHELL := env PATH=$(PATH) /bin/bash
