重新导入 soomla 包后出现此错误。如何解决这个问题?
重复符号 _NewBase64Encode_soomla 在:/Users/Nikunj/Unity/2D/NoOneCanDoIt/Build/b2/Libraries/Plugins/iOS/Soomla/libSoomlaiOSCore.a(NSData-Base64.o) /Users/Nikunj/Unity/2D/NoOneCanDoIt/Build /b2/Libraries/Plugins/iOS/SoomlaShared/libKeeva.a(NSData-Base64.o) 重复符号 _NewBase64Decode_soomla 在:/Users/Nikunj/Unity/2D/NoOneCanDoIt/Build/b2/Libraries/Plugins/iOS/Soomla/libSoomlaiOSCore .a(NSData-Base64.o) /Users/Nikunj/Unity/2D/NoOneCanDoIt/Build/b2/Libraries/Plugins/iOS/SoomlaShared/libKeeva.a(NSData-Base64.o) 重复符号 _OBJC_CLASS_$_FBEncryptorAES 在:/用户/Nikunj/Unity/2D/NoOneCanDoIt/Build/b2/Libraries/Plugins/iOS/Soomla/libSoomlaiOSCore.a(FBEncryptorAES.o) /Users/Nikunj/Unity/2D/NoOneCanDoIt/Build/b2/Libraries/Plugins/iOS /SoomlaShared/libKeeva.a(FBEncryptorAES.o) 重复符号 _OBJC_METACLASS_$_FBEncryptorAES 在:/Users/Nikunj/Unity/2D/NoOneCanDoIt/Build/b2/Libraries/Plugins/iOS/Soomla/libSoomlaiOSCore.a(FBEncryptorAES.o) /Users/Nikunj/Unity/2D/NoOneCanDoIt /Build/b2/Libraries/Plugins/iOS/SoomlaShared/libKeeva.a(FBEncryptorAES.o) ld:架构 armv7 clang 的 4 个重复符号:错误:链接器命令失败,退出代码为 1(使用 -v 查看调用) 首先显示仅 200 条警告