'How to debug any file in node JS without specifying the file name in launch,json

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I want to debug both workout.js and test.js but vscode is debugging only the workout.js how do I set the file name dynamic. so I can debug any js file with one launch.json config

Like "program": "${workspaceFolder}/${currentFileName}.js",

Solution 1:^[1]

Take a look at the official documentation about variable substitution in Debugging and Task configuration files, there are a few that can be used to achieve what you want.

Solution 2:^[2]

The easy way is to add multiple settings inside your single lunch config file. and whenever you want to debug any then choose that name while debugging

Solution 3:^[3]

Reffer this for more details Documnetaion

{
    "version": "0.2.0",
    "configurations": [
        {
            "name": "Node js",
            "program": "${file}",
            "request": "launch",
            "skipFiles": ["<node_internals>/**"],
            "type": "node"
        }
    ]
}

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Source: Stack Overflow

Solution	Source
Solution 1	Bernat
Solution 2	Kishan Vaishnani
Solution 3	Raghavan Vidhyasagar

'How to debug any file in node JS without specifying the file name in launch,json

Solution 1:[1]

Solution 2:[2]

Solution 3:[3]

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Solution 1:^[1]

Solution 2:^[2]

Solution 3:^[3]