xcodebuild - libiconv编译错误

Question

我正在为一个iOS项目制作一个构建脚本。该项目是通过XCode构建的；但是，使用xcodebuild命令，我得到了一个奇怪的链接错误：

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)

搜索谷歌等还没有透露出解决方案。

libiconv.dylib列在“链接二进制”er框架中。

我完全搞不懂这件事。就像该项目通过XCode本身进行编译一样

下面是我用来启动构建的命令：

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

Answer 1

我发现要修复这个问题，必须将$(继承)添加到Library搜索路径中。这应该搜索iOS SDK中的libiconv，而不是计算机上共享的动态库。