我已经习惯了一段时间。当我必须运行小脚本或者似乎与构建阶段无关的东西时(例如,调试,打包,构建外部库等等),我只是使用Makefile项目,如下所示
-- [[ X. External Makefile libraries ]]
project "external"
kind "Makefile"
location "build"
buildcommands {
"cd ..; make -f Makefile"
}
cleancommands {
"cd ..; make -f Makefile clean"
}
-- [[ X+1. Integration ]]
project "integration"
kind "Makefile"
location "build"
buildcommands {
-- PacketNgin Application Library
"ar x ../libc.a",
"ar x ../libm.a",
"ar x ../libtlsf.a",
"ar x ../libcore.a",
"ar x ../libexpat.a",
"ar rcs ../libpacketngin.a *.o",
"cp -rL ../core/include/* ../../include",
"cp -rL ../expat/include/* ../../include",
"cp -rL ../openssl/include/* ../../include",
"cp -rL ../zlib/*.h ../../include",
"rm ./*.o -rf",
-- Linux Application library
"ar x ../libtlsf.a ", -- Blank is added at the end on purpose
"ar x ../libcore_linux.a",
"ar rcs ../libumpn.a *.o",
"rm ./*.o -rf ", -- Blank is added at the end on purpose
}
cleancommands {
"rm *.o -rf",
"rm ../*.a -rf"
}
我意识到这种做法非常混乱,因为它不会将真正的构建Makefile与虚假目标分开,甚至可以构建不必要的Makefile。所以,我想弄清楚通过premake生成假目标。
我考虑了newaction语法,但我发现它只是为premake脚本而不是Makefile目标制作目标。
是否有通过预制生成虚假目标的最佳实践或方法?
答案 0 :(得分:0)
Premake目前不支持创建任意虚假目标(尽管您可以submit a feature request或自己处理并create a pull request)。
但可以使用Premake本身为您运行命令。这是一个简单的例子,它创建了一个名为" integration":
的新动作i = 2
For Each c In Range("D2", Range("D" & Rows.Count).End(xlUp))
CB_Account.AddItem Cells(i, 3).Value
CB_Account.AddItem Cells(i, 4).Value
i = i + 1
Next
添加到项目脚本后,您可以将其命名为:
function executeAll(commands)
for _, command in ipairs(commands) do
os.execute(command)
end
end
newaction
{
trigger = "integrate",
description = "Run integration steps",
execute = function ()
executeAll {
"ar x ../libc.a",
"ar x ../libm.a",
"ar x ../libtlsf.a",
"ar x ../libcore.a",
"ar x ../libexpat.a",
"ar rcs ../libpacketngin.a *.o",
-- and so on...
}
end
}