xcode - 将 armv6 架构添加到使用 Xcode 4.5 错误构建的 iOS 可执行文件的脚本

Question

你能帮我一点吗？

如果我使用以下脚本：

# Find the common base directory for both build
XCODE_BUILD=${BUILD_ROOT%%/Build*}
# Change this to the full path where Xcode 4.4 (or below) puts your armv6 output, using the previously derived base
export ARMV6_EXECUTABLE_PATH="$XCODE_BUILD/Build/Products/Release_armv6-iphoneos/$EXECUTABLE_PATH"

然后构建失败，消息如下：

Missing }.
Command /bin/sh failed with exit code 1

我做错了什么？

谢谢你的帮助。

Answer 1

我解决了我的错误。

因此，如果您从下一行更改上述行，您的应用程序可以构建，也可以存档。

# Find the common base directory for both build
setenv XCODE_BUILD `echo $BUILD_ROOT| sed 's|Build.*||'`
echo $XCODE_BUILD
# Change this to the full path where Xcode 4.4 (or below) puts your armv6 output, using the previously derived base
setenv ARMV6_EXECUTABLE_PATH "$XCODE_BUILD/Build/Products/Release_armv6-iphoneos/$EXECUTABLE_PATH"