Issue
I'm trying to set up another Xcode project as the one passed on launch for the purposes of testing an Xcode extension:
Rather than hardcoding a path like this, I'd like to make it relative to the current directory.
I created a placeholder build script and output the environment variables. I found out that the current project directory can be accessed via the PROJECT_DIR
environment variable.
How do I reference this environment variable in the Arguments Passed On Launch screen?
For example, instead of /Users/zoesmith/Desktop/Sample/TestableCode.xcodeproj
, I'd like the path to be something like this instead:
PROJECT_DIR/../TestableCode.xcodeproj
Solution
Looks like the following works:
"$(PROJECT_DIR)/../TestableCode.xcodeproj"
I think the quotes are required in case PROJECT_DIR
has any spaces in it.
I'm not sure what the difference is between $PROJECT_DIR
and $(PROJECT_DIR)
, but both seem to work.
Answered By - Senseful Answer Checked By - Mary Flores (PHPFixing Volunteer)
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