使用xcbuild / xcrun和xcconfig为cordova ios应用程序添加64位支持

Question

我正在使用cordova应用程序，我设法使用脚本从命令行创建ipa。

echo "" > $PROJECT_DIR/cordova/build.xcconfig
echo "CODE_SIGN_IDENTITY = iPhone Distribution: XXXXXXX" >> $PROJECT_DIR/cordova/build.xcconfig
echo "CODE_SIGN_RESOURCE_RULES_PATH = \$(SDKROOT)/ResourceRules.plist" >> $PROJECT_DIR/cordova/build.xcconfig
echo "IPHONEOS_DEPLOYMENT_TARGET = 7.0" >> $PROJECT_DIR/cordova/build.xcconfig

cordova build ios --device

xcrun \
  -sdk iphoneos PackageApplication \
  -v "$PRODUCT_DIR/$TARGET.app" \
  -o "$OUTDIR/$TARGET.ipa" \
  --embed "$PROVISONING_PROFILE" \
  --sign "$IDENTITY"

这很好用，但是当我上传到Apple TestFlight时，我收到此警告。

你们对此有任何暗示吗？也许我可以在.xcconfig添加一行似乎非常方便！

Answer 1

我想错了！ --release标志丢失了。现在我可以在不打开XCode的情况下构建！

echo "\n > patching cordova xconfig for distribution\n"
echo "" > $PROJECT_DIR/cordova/build.xcconfig
echo "CODE_SIGN_IDENTITY = $IDENTITY" >> $PROJECT_DIR/cordova/build.xcconfig
echo "CODE_SIGN_RESOURCE_RULES_PATH = \$(SDKROOT)/ResourceRules.plist" >> $PROJECT_DIR/cordova/build.xcconfig
echo "IPHONEOS_DEPLOYMENT_TARGET = 7.0" >> $PROJECT_DIR/cordova/build.xcconfig

echo "\n > cleanup $OUTDIR/$TARGET.ipa\n"
rm -vf $OUTDIR/$TARGET.ipa

echo "\n > build project\n"
cordova build ios --release --device

echo "\n > creating ipa at $OUTDIR/$TARGET.ipa\n"
cd $BASE_DIR
xcrun \
  -sdk iphoneos PackageApplication \
  -v "$PRODUCT_DIR/$TARGET.app" \
  -o "$OUTDIR/$TARGET.ipa" \
  --embed "$PROVISONING_PROFILE" \
  --sign "$IDENTITY"