xcodebuild - libiconv - 编译错误

我正在为iOS项目制作一个构建脚本。该项目通过XCode构建;然而，与xcodebuild联编命令我得到一个奇怪的链接错误：xcodebuild - libiconv - 编译错误

ld: warning: ignoring file /sw/lib//libiconv.dylib, file was built for x86_64 which is not the architecture being linked (armv7): /sw/lib//libiconv.dylib 
Undefined symbols for architecture armv7: 
"_iconv_open", referenced from: 
    l2451 in libscanditsdk-iphone-3.1.1.a(mirasense.o) 
"_iconv", referenced from: 
    l2451 in libscanditsdk-iphone-3.1.1.a(mirasense.o) 
"_iconv_close", referenced from: 
    l2451 in libscanditsdk-iphone-3.1.1.a(mirasense.o) 
ld: symbol(s) not found for architecture armv7 
clang: error: linker command failed with exit code 1 (use -v to see invocation)

搜索谷歌和SO没有透露解决方案。

libiconv.dylib列在“链接的二进制”呃框架的东西。

我完全沉迷于这一个。像项目并通过了XCode本身

编译下面是我使用启动构建命令：

xcodebuild -target "${TARGET_NAME}" -sdk "${TARGET_SDK}" -configuration Release -scheme "${SCHEME_NAME}" PROVISIONING_PROFILE="${PROJ_PROF_UUID}"

来源

2013-10-04 TMacGyver

我发现，要解决这个问题，你有加$（继承）的库搜索路径。这应该搜索iOS SDK中的libiconv，而不是在计算机上共享的动态库

来源

2014-01-03 19:29:04 TMacGyver

xcodebuild - libiconv - 编译错误

回答

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