ios - Quickblox:体系结构 x86_64 的 undefined symbol :错误

Question

我使用了以下 Pod，但出现错误

Undefined symbols for architecture x86_64:

platform :ios, "8.0" use_frameworks!

target 'CSocial' do

#
pod 'Alamofire', '~> 3.0'

pod 'FBSDKCoreKit'

pod 'FBSDKLoginKit'

pod 'FBSDKShareKit'

pod 'FlickrKit'

pod 'MONActivityIndicatorView'

聊天集成

pod 'QuickBlox'
#Production pod
#pod 'QMCVDevelopment', :git => 'git@github.com:QuickBlox/QMChatViewController-ios.git', :commit => '7244eaf609797a5dc5ec6d28dc2b284633b03a1f'

#For development pods
pod 'QMCVDevelopment', :git => 'https://github.com/QuickBlox/QMChatViewController-ios.git'

#Production pod
#pod 'QMServicesDevelopment', :git => 'git@github.com:QuickBlox/q-municate-services-ios.git', :commit => 'ab56ac143bb770e76409a027bfb38de582d728aa'

#For development pods
pod 'QMServicesDevelopment', :git => 'https://github.com/QuickBlox/q-municate-services-ios.git'

结束

Answer 1

执行以下步骤即可解决您的问题:

1) 点击项目中的“Pods.xcodeproj”

2) 选择所有 Pod/您遇到问题的任何一个 Pod。

3)点击“build设置”并将“仅构建事件架构”更改为“否”

4)现在尝试构建来构建项目，您的问题将得到解决。希望这有帮助!