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Accepted
felix0000
Asked: 2020-06-28 10:20:22 +0800 CST 2020-06-28 10:20:22 +0800 CST

Use Child process spawn in Node JS and save data in variable

  • 772

The idea is to execute a command and capture the response in a variable so that it can be processed.

const { stdout } = require('process');

function files(command, arg) {
    const { spawn } = require('child_process');
    const ls = spawn(command, [arg, '/home']);
    ls.stdout.on('data', (data) => {
        console.log(`stdout: ${data}`);
        return data;
    });
}

var ej = files("dir", "")
console.log("eje => " + ej);

When executing the following line you get:

 console.log(`stdout: ${data}`);

    stdout: /home/test/Descargas: a.jpg  link.txt

When trying to return the value appears undefined

return data; 
console.log("eje => " + ej);

eje=> undefined
javascript

1 Answers

  1. Best Answer

    Code with fixes:

     const { spawnSync } = require('child_process');
    
    function files(command, arg, path) {
        const ls = spawnSync(command, [arg, path], { encoding: 'utf8' });
        return ls.stdout;
    };
    
    
    var ej = files("dir", "", '/home/test/Descargas');
    console.log("eje => " + ej);

    Departure

    eje => /home/test/Descargas: a.jpg  link.txt
    • 0