Ios 如何在xcode6中创建通用框架

我知道如何在Xcode 5中创建框架。但是在Xcode 6中，如何将模拟器框架和设备框架结合起来呢？当我尝试合并时，会出现代码签名错误。当我使用

lipo

组合这两个框架时，我也会得到一个错误

---

错误：

命令/bin/sh失败，退出代码为65

我在xcode6中创建通用框架的解决方案

尝试以下步骤：

第1步：

File—> 
    New —> 
       Project —> 
           Framework & Library —> 
               Next —> 
                      Product Name

File ->
    New ->
           Target ->
               iOS ->
                      Other -> 
                          Aggrigate ->somename eg: framework

From Targets —> 
    Custom aggregate target(Eg: Framework)—> 
           Build Phase—> 
               Add Run script

Goto active scheme —> 
    Custom aggregate —> 
           Build

步骤2:创建自定义类文件

步骤3：

Target -> 
       Build phase -> 
           Headers, 

使所有头文件成为公共文件。现在内置模拟器和设备

第4步：

File—> 
    New —> 
       Project —> 
           Framework & Library —> 
               Next —> 
                      Product Name

File ->
    New ->
           Target ->
               iOS ->
                      Other -> 
                          Aggrigate ->somename eg: framework

From Targets —> 
    Custom aggregate target(Eg: Framework)—> 
           Build Phase—> 
               Add Run script

Goto active scheme —> 
    Custom aggregate —> 
           Build

第五步：

File—> 
    New —> 
       Project —> 
           Framework & Library —> 
               Next —> 
                      Product Name

File ->
    New ->
           Target ->
               iOS ->
                      Other -> 
                          Aggrigate ->somename eg: framework

From Targets —> 
    Custom aggregate target(Eg: Framework)—> 
           Build Phase—> 
               Add Run script

Goto active scheme —> 
    Custom aggregate —> 
           Build

步骤6:在运行脚本中添加以下

shell代码

///

UNIVERSAL_OUTPUTFOLDER=${BUILD_DIR}/${CONFIGURATION}-universal

# make sure the output directory exists
mkdir -p "${UNIVERSAL_OUTPUTFOLDER}"

# Step 1. Build Device and Simulator versions
xcodebuild -target "${PROJECT_NAME}" ONLY_ACTIVE_ARCH=NO -configuration ${CONFIGURATION} -sdk iphoneos  BUILD_DIR="${BUILD_DIR}" BUILD_ROOT="${BUILD_ROOT}" clean build
xcodebuild -target "${PROJECT_NAME}" -configuration ${CONFIGURATION} -sdk iphonesimulator -arch x86_64 BUILD_DIR="${BUILD_DIR}" BUILD_ROOT="${BUILD_ROOT}" clean build

# Step 2. Copy the framework structure to the universal folder
cp -R "${BUILD_DIR}/${CONFIGURATION}-iphoneos/${PROJECT_NAME}.framework" "${UNIVERSAL_OUTPUTFOLDER}/"

# Step 3. Create universal binary file using lipo and place the combined executable in the copied framework directory
lipo -create -output "${UNIVERSAL_OUTPUTFOLDER}/${PROJECT_NAME}.framework/${PROJECT_NAME}" "${BUILD_DIR}/${CONFIGURATION}-iphonesimulator/${PROJECT_NAME}.framework/${PROJECT_NAME}" "${BUILD_DIR}/${CONFIGURATION}-iphoneos/${PROJECT_NAME}.framework/${PROJECT_NAME}"

第7步：

File—> 
    New —> 
       Project —> 
           Framework & Library —> 
               Next —> 
                      Product Name

File ->
    New ->
           Target ->
               iOS ->
                      Other -> 
                          Aggrigate ->somename eg: framework

From Targets —> 
    Custom aggregate target(Eg: Framework)—> 
           Build Phase—> 
               Add Run script

Goto active scheme —> 
    Custom aggregate —> 
           Build

第8步：现在右键单击Xcode中产品的框架，然后单击finder中的显示

选中“调试通用”文件夹并获取通用框架。
替换

xcodebuild -target "${PROJECT_NAME}" -configuration ${CONFIGURATION} -sdk iphonesimulator -arch x86_64 BUILD_DIR="${BUILD_DIR}" BUILD_ROOT="${BUILD_ROOT}" clean build

与

clean build
您正在创建静态库？您的联合库（.a文件）意味着（debug.a&release.a）一旦向我显示您的lipo语句，这对我来说非常有效。我所做的唯一更改是删除了-arch x86_64。这导致了无法在i386模拟器上运行的问题（5s之前的任何内容）.-sdk iphonesimucator为编译器提供了足够的信息，可以为i386和x86_64生成。如果您明确指定其中一个，它将排除另一个。谢谢！@Chad您是如何删除的-arch x86_64。因为当我从运行脚本中删除时，它仍然没有为模拟器运行。@Chad：您需要尝试步骤8。然后，只有您才能获得通用框架支持设备和模拟器。@Chad我也不在模拟器中运行。@AnsariAwais你能让我设置部署目标的意思吗？你做了什么？