osx mavericks - How to select XCode 3.2.6 with xcode-select? -
i have upgraded mac os 10.9.3 mac os 10.6.8. had xcode 3.2.6 installed. new developments, needed install xcode 4.5.2.
in maven build, invoke xcodebuild build c++ projects. problem need xcodebuild 3.2.6 build projects, , xcodebuidl 4.5.2 new ones.
i need able choose on-the-fly xcodebuild version launch build on project. saw xcode-select me doing this, stuck on way use xcode-select choose xcodebuild 3.2.6.
considerong xcode 4.5.2 installed @ /applications/xcode.app, using command
sudo xcode-select -switch /applications/xcode.app
does job, as
xcodebuild -version
outputs
xcode 4.5.2 build version 4g2008a
considering xcode 3.2.6 @ /developer/applications/xcode.app, problem is, why command not work?
sudo xcode-select -switch /developer/applications/xcode.app
it outputs
xcode-select: error: invalid developer directory '/developer/applications/xcode.app'
the workaround find using absolute path use xcodebuild 3.2.6, that's tricky:
/developer/usr/bin/xcodebuild
is there clean way select xcodebuild 3.2.6 xcode-select?
you have point path /contents/developer
folder in xcode.app
bundle.
try this:
$ sudo xcode-select -switch /developer/applications/xcode.app/contents/developer
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