Question

我正在为iOS项目制作构建脚本。该项目通过XCode构建;但是，使用xcodebuild命令，我得到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Google和SO尚未透露解决方案。

libiconv.dylib列在“链接二进制”框架之下。

我对这个完全感到难过。就像项目通过XCode本身编译一样

这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

我发现要解决此问题，您已将$（继承）添加到库搜索路径。这应该在iOS SDK中搜索libiconv而不是在计算机上共享的动态库

xcodebuild - libiconv - 编译错误

1 个答案: