what's an ideal way to get path of a caller file

i've found the method that works for me, but it's too hacky to be true lmao

function() {
    let callerPath = new Error().stack!.split("\n")[2]!;
    for (let i = 2; i > 0; i--) {
        callerPath = callerPath.slice(
            callerPath.lastIndexOf("("),
            callerPath.lastIndexOf(":")
        );
    }
    return callerPath.slice(1, callerPath.length);
}

function() {
    let callerPath = new Error().stack!.split("\n")[2]!;
    for (let i = 2; i > 0; i--) {
        callerPath = callerPath.slice(
            callerPath.lastIndexOf("("),
            callerPath.lastIndexOf(":")
        );
    }
    return callerPath.slice(1, callerPath.length);
}

so i'm asking, is there no better way to do the same thing?

Solution:

whatever, i use process.cwd()

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2 Replies

Solution

🐀•2mo ago

whatever, i use process.cwd()

🐀OP•2mo ago

nah, this idea can be evolved and improved. but i've found flaws already, and thank you i don't have to be that dependent on a stack

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what's an ideal way to get path of a caller file

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