Set the CONFIGURATION_BUILD_DIR Build Setting to be the directory you want.
So in your example:
xcodebuild ...cut... CONFIGURATION_BUILD_DIR=<desired dir> ...cut...
Question
Is there a way I can choose the destination folder (directory) for when I build an iOS app via command line using xcodebuild?
Right now I am using a command like
xcodebuild -sdk iphonesimulator -workspace /<path_to_project>/ios-app/CompanyName-iOS.xcworkspace -scheme AppName -configuration Debug RUN_APPLICATION_TESTS_WITH_IOS_SIM=YES ONLY_ACTIVE_ARCH=NO clean build 2>&1
In the output I see something like
GenerateDSYMFile /Users/<username>/Library/Developer/Xcode/DerivedData/CompanyName-iOS-fcvhojqctgtmvgdaavahmjutbfyy/Build/Products/Debug-iphonesimulator/AppNAme.app.dSYM
Is there a way I could get this location where the app is output to or even specify where I would rather have the app be sent to? I need access to the app to run Appium selenium tests and just having a build execute and run tests is not helpful. Also trying to incorporate jenkins into the mix and need to have commands to fully automate the process
La solution
Set the CONFIGURATION_BUILD_DIR Build Setting to be the directory you want.
So in your example:
xcodebuild ...cut... CONFIGURATION_BUILD_DIR=<desired dir> ...cut...
Autres conseils
use -derivedDataPath path to get compiled output onto specific folder.
xcodebuild -scheme xxx -workspace xxx.xcworkspace ONLY_ACTIVE_ARCH=NO -sdk iphonesimulator -configuration Debug -derivedDataPath ./build build