我需要为iPhone创建一个应用程序，它将使用libpq连接到PostgreSQL 8.4数据库。问题是我无法获得一个链接到libpq进行编译的简单iPhone。然而，我能够获得相当于常规Mac桌面应用程序的应用程序来编译和连接到PostgreSQL而没有任何问题。我在Snow Leopard上运行Xcode 3.2。
我正在为arm和x86_84构建libpq。 arm构建适用于真正的iPhone，x86_64适用于iPhone模拟器。然后我创建一个包含两个文件的胖二进制文件，最后得到一个名为libpq的文件。这个文件是我在常规Mac应用程序中使用的文件，它可以正常工作，并在尝试构建iPhone应用程序时导致问题。
这是构建libpq时的构建脚本。
#!/bin/bash
DEVROOT=/Developer/Platforms/iPhoneOS.platform/Developer
SDKROOT=$DEVROOT/SDKs/iPhoneOS3.0.sdk
rm -rf /Users/bob/mylibs
mkdir /Users/bob/mylibs #Store there compiled libs
make clean
#Build ARM library
./configure --host=arm-apple-darwin --without-readline --disable-ipv6 CC=$DEVROOT/usr/bin/arm-apple-darwin9-gcc-4.0.1 CPPFLAGS="-I$SDKROOT/usr/lib/gcc/arm-apple-darwin9/4.0.1/include/ -I$SDKROOT/usr/include/" CFLAGS="$CPPFLAGS -arch armv6 -pipe -no-cpp-precomp -isysroot $SDKROOT" CPP="$DEVROOT/usr/bin/cpp $CPPFLAGS" LD=$DEVROOT/usr/bin/ld
make -C src/interfaces/libpq
cp /Users/bob/Downloads/postgresql-8.4.1/src/interfaces/libpq/libpq.a /Users/bob/mylibs/libpq.arm
#Then build i386 library
make clean && ./configure && make -C src/interfaces/libpq
cp src/interfaces/libpq/libpq.a /Users/bob/mylibs/libpq.i386
#Then make fat binary
$DEVROOT/usr/bin/lipo -arch armv6 /Users/bob/mylibs/libpq.arm -arch x86_64 /Users/bob/mylibs/libpq.i386 -create -output /Users/bob/mylibs/libpq
当我尝试从Xcode中编译iPhone应用程序时，这是构建日志。
Build iPhonePg of project iPhonePg with configuration Debug
Ld build/Debug-iphonesimulator/iPhonePg.app/iPhonePg normal i386
cd /Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg
setenv MACOSX_DEPLOYMENT_TARGET 10.5
setenv PATH "/Developer/Platforms/iPhoneSimulator.platform/Developer/usr/bin:/Developer/usr/bin:/usr/bin:/bin:/usr/sbin:/sbin"
/Developer/Platforms/iPhoneSimulator.platform/Developer/usr/bin/gcc-4.2 -arch i386 -isysroot /Developer/Platforms/iPhoneSimulator.platform/Developer/SDKs/iPhoneSimulator3.0.sdk -L/Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg/build/Debug-iphonesimulator -L../../../../mylibs -L/Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg -L/Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg/../../../../mylibs -F/Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg/build/Debug-iphonesimulator -filelist /Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg/build/iPhonePg.build/Debug-iphonesimulator/iPhonePg.build/Objects-normal/i386/iPhonePg.LinkFileList -mmacosx-version-min=10.5 -framework Foundation -framework UIKit -framework CoreGraphics /Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg/libpq -o /Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg/build/Debug-iphonesimulator/iPhonePg.app/iPhonePg
ld: warning: in /Users/bob/Documents/Programming/PragProgrammerIphoneSDK/iPhonePg/libpq, missing required architecture i386 in file
Undefined symbols:
"_PQclear", referenced from:
-[iPhonePgAppDelegate applicationDidFinishLaunching:] in iPhonePgAppDelegate.o
"_PQerrorMessage", referenced from:
-[iPhonePgAppDelegate applicationDidFinishLaunching:] in iPhonePgAppDelegate.o
"_PQconnectdb", referenced from:
-[iPhonePgAppDelegate applicationDidFinishLaunching:] in iPhonePgAppDelegate.o
"_PQfinish", referenced from:
-[iPhonePgAppDelegate applicationDidFinishLaunching:] in iPhonePgAppDelegate.o
"_PQstatus", referenced from:
-[iPhonePgAppDelegate applicationDidFinishLaunching:] in iPhonePgAppDelegate.o
"_PQexec", referenced from:
-[iPhonePgAppDelegate applicationDidFinishLaunching:] in iPhonePgAppDelegate.o
ld: symbol(s) not found
collect2: ld returned 1 exit status
其他人遇到这个可以帮忙吗？
由于 StartShip3000
答案 0 :(得分：5)
看起来你实际上并没有链接到库中。您已将其所在目录添加到搜索路径中，但我未在链接行的任何位置看到-lpq。
此外，模拟器环境仅为32位，x86_64库不适用于模拟器二进制文件。
答案 1 :(得分：1)
好的，谢谢路易斯的评论。我通过更改构建i386版本的makefile行来完成此操作。我添加了CFLAGS = -arch i386“看起来像这样
#Then build i386 library
make clean && ./configure CFLAGS="-arch i386" && make -C src/interfaces/libpq
我现在有另一个问题，我将在此链接中引用一个新问题。
答案 2 :(得分：1)