QT PRI文件中的文件路径通过函数

发布于 2025-02-05 01:36:28 字数 822 浏览 5 评论 0原文

我正在.pri文件中创建一个函数，该函数将包含在一个.pro文件中，目的是能够执行：

template(subdirs)
addMyProject(libA)
addMyProject(libB)
addMyProject(libC)

而不是

template(subdirs)
SUBDIRS += libA
libA.path = source/
SUBDIRS += libB
libB.path = source/
SUBDIRS += libC
libC.path = source/

如果我想使用依赖关系关键字，

同一问题。我当前的实现是以下一个：

defineTest(addMyProject) {

  library_name = $$1

  SUBDIRS += $$library_name
  export(SUBDIRS)

  {$$library_name}.file = {$$library_name}/source/{$$library_name}.pro
  QMAKE_EXTRA_TARGETS += {$$library_name}
  export(QMAKE_EXTRA_TARGETS)
}

子段部分还可以。我试图以不同的方式使其适用于.file部分，但我仍然没有找到解决方案。

有提示吗？

[编辑]

每个库都包含不同的子项目，目标是让QT创建者仅查看其中一个（路径源/LibraryName.pro的一个）。问题在于所有库都正确显示，但也是所有其他子项目。基本上，只有subdirs变量才能识别，但不识别.file关键字）

原文

I am creating a function in a .pri file that will be included in a .pro file with the intent of being able to do:

template(subdirs)
addMyProject(libA)
addMyProject(libB)
addMyProject(libC)

instead of

template(subdirs)
SUBDIRS += libA
libA.path = source/
SUBDIRS += libB
libB.path = source/
SUBDIRS += libC
libC.path = source/

Same problem if I want to use the depend keyword.

My current implementation is the following one:

defineTest(addMyProject) {

  library_name = $1

  SUBDIRS += $library_name
  export(SUBDIRS)

  {$library_name}.file = {$library_name}/source/{$library_name}.pro
  QMAKE_EXTRA_TARGETS += {$library_name}
  export(QMAKE_EXTRA_TARGETS)
}

The SUBDIRS part is ok. I have tried to make it work for the .file part in different ways, but still I have not found a solution.

Any hints?

[EDIT]

Each library contains different sub-projects, and the goal is to let Qt creator see only one of them (the one at the path source/libraryname.pro).
The issue is that all the libraries are correctly displayed, but also all the other sub-projects. Basically, only the SUBDIRS variable is recognised, but not the .file keyword)

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