打包命令行输出IPA文件

来源：互联网发布：大学生网络课程答案编辑：程序博客网时间：2024/05/28 11:29
//—————————————————ipa 成功了－－－－－－－－－－－－－－－－－－／# 工程名APP_NAME="1"# 目标名TARGET_NAME="1"# 证书CODE_SIGN_DISTRIBUTION="iPhone Distribution: 证书"PROVISIONING_PROFILE="1"# info.plist路径project_infoplist_path="./${APP_NAME}/info.plist"#取版本号bundleShortVersion=$(/usr/libexec/PlistBuddy -c "print CFBundleShortVersionString" "${project_infoplist_path}")#取build值bundleVersion=$(/usr/libexec/PlistBuddy -c "print CFBundleVersion" "${project_infoplist_path}")#DATE="$(date +%Y%m%d)"#IPANAME="${APP_NAME}_V${bundleShortVersion}_${DATE}.ipa"#ipa、dSYM文件DATE="$(date +%H%M%S)"IPA_NAME="${TARGET_NAME}_build${bundleVersion}_${DATE}.ipa"ARCH_NAME="${TARGET_NAME}_build${bundleVersion}_${DATE}.xcarchive"ZIP_NAME="${TARGET_NAME}_build${bundleVersion}_${DATE}.zip"echo "================= #本地存放ipa、dSYM路径 ${IPA_NAME}================="#本地存放ipa、dSYM路径LOCAL_IPA_PATH="output/${IPA_NAME}"LOCAL_ARCH_PATH="output/${ARCH_NAME}"LOCAL_ZIP_PATH="output/${ZIP_NAME}"#创建路径rm -rf "output"mkdir -p "output"echo "================= Xcode Clean ================="xcodebuild -target "${APP_NAME}.xcodeproj" -scheme "工程名" -configuration 'Release' cleanecho "================= Xcode Archive 开始================="xcodebuild archive -target "${APP_NAME}.xcodeproj" -scheme "工程名" -archivePath "${LOCAL_ARCH_PATH}" -configuration 'Release'echo "================= Xcode Archive success ================="xcodebuild  -exportArchive -exportFormat IPA -archivePath "${LOCAL_ARCH_PATH}" -exportPath "${LOCAL_IPA_PATH}" -exportProvisioningProfile "${PROVISIONING_PROFILE}"zip -r ${LOCAL_ZIP_PATH} ${LOCAL_ARCH_PATH}//—————————————————ipa 成功了－－－－－－－－－－－－－－－－－－／
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