我有两个cmake unit-test来检查url和json助手类，在我的项目的CMakeLists.txt中声明如下：

########################################
# CMakeLists.txt - Tests               #
########################################
add_executable(urltest test/url.cpp)
add_test(NAME URLTest COMMAND urltest)

add_executable(jsontest test/json.cpp)
add_test(NAME JSONTest COMMAND jsontest )

我在VSCode的testing视图中直观地得到了测试单元，并且我可以在发布模式下完美地运行这两个单元：Execute the unit tests in release mode
现在，如果我想用另一个按钮调试它，我必须选择一个启动配置，为每个单元测试手动创建一个启动配置需要什么：Execute the unit tests in debug mode
我目前的launch.json文件要求每个单元测试都有一个配置，这是不可能维护的：

{ // launch.json
    "version": "0.2.0",
    "configurations": [
        {
            "name": "CTest-urltest",
            "type": "lldb",
            "request": "launch",
            "program": "${workspaceFolder}/build/urltest",
            "args": [],
            "initCommands": ["breakpoint set -n main -N entry"],
            "exitCommands": ["breakpoint delete entry"],
            "cwd": "${workspaceFolder}"
        },
        {
            "name": "CTest-jsontest",
            "type": "lldb",
            "request": "launch",
            "program": "${workspaceFolder}/build/jsontest",
            "args": [],
            "initCommands": ["breakpoint set -n main -N entry"],
            "exitCommands": ["breakpoint delete entry"],
            "cwd": "${workspaceFolder}"
        }
    ]
}

我猜问题只出在program属性上，如何将VSCode的testing视图上点击的程序Map到唯一的启动配置？例如：

{ // launch.json
    "version": "0.2.0",
    "configurations": [
        {
            "name": "CTest",
            "type": "lldb",
            "request": "launch",
            "program": "${workspaceFolder}/build/${command: testing.currentPath}",
            "args": [],
            "initCommands": ["breakpoint set -n main -N entry"],
            "exitCommands": ["breakpoint delete entry"],
            "cwd": "${workspaceFolder}"
        }
    ]
}