如何从 Applescript 启动特定版本的 Xcode? [英] How do I launch a particular version of Xcode from Applescript?

问题描述

我安装了两个版本的 Xcode，Xcode 3.2.3 和 Xcode4 开发者预览版.我如何确保从 Applescript 中选择了 3.2.3 版本?

I have two versions of Xcode installed, Xcode 3.2.3 and the Xcode4 developer preview. How do I ensure from Applescript that the 3.2.3 version is picked?

推荐答案

不是简单地通过名称引用 Xcode，即:

Instead of simply referencing Xcode by its name, i.e.:

tell application "Xcode"
    ...
end tell

您还可以通过完整的 POSIX 路径引用应用程序的特定版本，即:

you can also reference a particular version of an application by its full POSIX path, i.e.:

tell application "/Developer/Applications/Xcode 3.2.3.app"
    ...
end tell

另请参阅 应用程序类.

更复杂的解决方案是在 Launch Services 数据库中搜索系统上安装的所有应用程序版本.然后，您可以以编程方式选择具有所需版本的那个:

A more complex solution involves searching the Launch Services database for all the versions of an application that are installed on the system. You can then programmatically pick the one with the required version:

property pLSRegisterPath : "/System/Library/Frameworks/CoreServices.framework/Versions/A/Frameworks/LaunchServices.framework/Versions/A/Support/lsregister"
property pAppBundleID : "com.apple.Xcode"
property pAppRequiredVersion : "3.2.3"

set theAppPaths to every paragraph of (do shell script pLSRegisterPath & " -dump | grep --before-context=2 \"" & pAppBundleID & "\" | grep --only-matching \"/.*\\.app\"")

set xcodeApp to missing value
repeat with theAppPath in theAppPaths
    try
        if (version of application theAppPath) = pAppRequiredVersion then
            set xcodeApp to application theAppPath
            exit repeat
        end if
    end try
end repeat

if xcodeApp = missing value then
    error "Needed application version not installed."
end if

using terms from application "Xcode"
    tell xcodeApp
        activate
    end tell
end using terms from

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