Question

我正在尝试为与iPhone模拟器应用程序相关的Xcode项目设置命令行参数。

当我尝试运行以下脚本时 “使用属性创建新的启动参数{name：”file：///Users/aakash/Desktop/sample_h.html“，active：yes}”

给出错误：执行错误：Xcode出错：无法将该元素制作或移动到该容器中。（-10024）

这是脚本：

!/bin/zsh

BUILD_PATH=$(dirname $0)

while [[ -z $BUILD_FILE && $BUILD_PATH != "/" ]]; do
    BUILD_FILE=$(find $BUILD_PATH -name '*.xcodeproj' -maxdepth 1)
    BUILD_PATH=$(dirname $BUILD_PATH)
done

if [[ -z $BUILD_FILE ]]; then
    echo "Couldn't find an xcode project file in directory"
    exit 1
fi

open -a Xcode "$BUILD_FILE"

BUILD_FILE=${BUILD_FILE//\//:}

SIMULATOR_SDKS=( /Developer/Platforms/iPhoneSimulator.platform/Developer/SDKs/*.sdk )

SIMULATOR_SDK=${SIMULATOR_SDKS[-1]} 
SIMULATOR_SDK_STRING=$(basename ${(L)SIMULATOR_SDK%.[a-z]*})

if [[ -z $SIMULATOR_SDK ]]; then
    echo "Couldn't find a simulator SDK"
    exit 1
fi

echo $BUILD_FILE
echo $BUILD_PATH

osascript <<SCRIPT
application "iPhone Simulator" quit
application "iPhone Simulator" activate


tell application "Xcode"
    open "$BUILD_FILE"
    set targetProject to project of active project document

    tell targetProject
        set active build configuration type to build configuration type "Debug"
        set active SDK to "$SIMULATOR_SDK_STRING"
        set value of build setting "SDKROOT" of build configuration "Debug" of active target to "$SIMULATOR_SDK"
        make new launch argument with properties{name:"file:///Users/aakash/Desktop/sample_h.html",active:no}

        if (build targetProject) is equal to "Build succeeded" then
                launch targetProject
        else
                application "iPhone Simulator" quit
        end if
    end tell
end tell
SCRIPT

任何线索??? 有没有其他方法为Xcode Project设置参数或我做错了吗？请帮忙。

Answer 1

对于从命令行构建，我通常使用Makefile启动xcodebuild，即Xcode的命令行前端。如果您愿意，也可以使用zsh脚本执行相同的操作。使用命令行工具设置项目构建选项非常简单。

Answer 2

您应该将活动可执行文件作为容器而不是Xcode应用程序

tell application "Xcode"
set targetProject to project of active project document
set targetExecutable to active executable of targetProject
tell targetExecutable
make new launch argument with properties {name:"new argument", active:true}
end tell
end tell

这仍适用于Xcode 3. *但不能长时间使用Xcode 4。*

如何通过AppleScript将命令行参数传递给Xcode项目？

2 个答案: