Return value from Node.JS event instead of console.log

Question

I have the following code which finds a file location for a given file. I want the value to be returned instead of console.log. Library used for finding file location - https://github.com/substack/node-findit

function searchfileloc(fname){
 var finder = require('findit')(__dirname)
 var path = require('path')

 finder.on('file', function (file) {
   if (path.basename(file) == fname){
       console.log(file); //return file
       finder.stop();
   }
  });

}

score 0 · Accepted Answer · answered Mar 10 '22 at 23:37

0

You can wrap the function in a Promise and call it using await to get the result.

Here's an example:

function searchfileloc(fname) {
  return new Promise((resolve, reject) => {
    const finder = require("findit")(__dirname);
    const path = require("path");

    let found = false;
    finder.on("file", (file) => {
      if (path.basename(file) === fname) {
        found = true;
        finder.stop();
        resolve(file);
      }
    });

    finder.on("end", () => {
      if (!found) {
        reject(new Error("File not found"));
      }
    });
  });
}

// Usage
const myFile = await searchfileloc("my_file_name");

answered Mar 10 '22 at 23:37

Ikbel

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It returns that await is not an async function. Does it require the wrapped function to be asynchronous as well? – Pranay Mishra Mar 11 '22 at 08:52
In that case the function def and function call has to inside the async block – Pranay Mishra Mar 11 '22 at 08:55
Yes, you'll need to use it inside an async function. – Ikbel Mar 11 '22 at 10:48

Return value from Node.JS event instead of console.log

1 Answers1