XCode 4调试器: 从命令行附加进程

Question

我想从命令行启动一个应用程序，并立即使用XCode 4中的图形调试器连接它，而无需点击Xcode菜单Product->Attach To Progress。是否可以使用bash或Apple脚本自动化这个过程？

- user2682950

Answer 1

这似乎在Xcode-Beta版本6.2（6C121）中有效。对于Xcode，只需将“Xcode-Beta”更改为Xcode。此外，您需要将进程名称“Staging”更改为要附加的进程。

tell application "Xcode-Beta"
activate
end tell

tell application "System Events"
    tell application process "Xcode"
        click (menu item "By Process Identifier (PID) or Name…" of menu 1 of menu item "Attach to Process" of menu 1 of menu bar item "Debug" of menu bar 1)
    end tell
    tell application process "Xcode"
        set value of text field 1 of sheet 1 of window 1 to "Staging"
    end tell
    tell application process "Xcode"
        click button "Attach" of sheet 1 of window 1
    end tell
end tell

- Canucklesandwich

Answer 2

适用于Xcode 12

tell application "Xcode"
    activate
end tell

tell application "System Events"
    tell application process "Xcode"
        click (menu item "Attach to Process by PID or Name…" of menu 1 of menu bar item "Debug" of menu bar 1)
        
    end tell
    tell application process "Xcode"
        set value of text field 1 of sheet 1 of window 1 to "Your App/Process Name"
    end tell
    tell application process "Xcode"
        click button "Attach" of sheet 1 of window 1
    end tell
end tell

tell application "our App/Process Name"
    activate
end tell

- season