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xcodecommand-linexcode4xcodebuild

How get the configs of a xcodeproj with command line (using xcodebuild)

I will create a build script with xcodebuild command.

But, I want get the config of my project, like PROJECT_DIR and use it for next action. Or better, I only want to know where the .app will be create.

Any suggest or I will have to read the internal xcodeproj files to get it?

Solution

  • Edit: In Product > Edit Scheme you can add a script to the "post-action" part of the dropdown menu, check "provide build settings from program" and export the $(BUILD_PRODUCTS_DIR) variable (or write it to a file, whatever).

    I did not get the meaning of your question at first, sorry.

    Old stuff: You can right-click on the target and select "show in Finder" to know where it is going to be built.

    If you don't like the location (default is a long path) you can change the preferences of Xcode to save the DerivedData-Folder directly in the project-folder. Then the path will be something like $(projectfolder)/DerivedData/Build/Product/program.

    I assume you are using Xcode4.