Saving the output of a child process in a variable in the parent in NodeJS

Question

I would like to start a child process in NodeJS and save it's output into a variable. The following code gives it to stdout:

require("child_process").execSync("echo Hello World", {"stdio": "inherit"});

I have something in mind that is similar to this code:

var test;
require("child_process").execSync("echo Hello World", {"stdio": "test"});
console.log(test);

The value of test was supposed to be Hello World.

Which does not work, since "test" is not a valid stdio value.

Perhaps this is possible using environment variables, however I did not find out how to modify them in the child process with the result still being visible to the parent.

Aurora0001 · Accepted Answer · 2016-12-06T17:52:01.167

20

execSync is a function which returns the stdout of the command you pass in, so you can store its output into a variable with the following code:

var child_process = require("child_process");
var test = child_process.execSync("echo Hello World");
console.log(test);
// => "Hello World"

Be aware that this will throw an error if the exit code of the process is non-zero. Also, note that you may need to use test.toString() as child_process.execSync can return a Buffer.

edited Dec 06 '16 at 17:52

answered Dec 06 '16 at 17:41

Aurora0001

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Thank you for your answer! I would like to add that I had to use `test.toString()`, otherwise the output was ``. – pfo Dec 06 '16 at 17:50
@pfo will note this - I think it varies with the version since my local copy didn't do that. – Aurora0001 Dec 06 '16 at 17:51
1

How do you do that if it exists with a non-zero exit code though? – fourk Jan 15 '21 at 17:08

Saving the output of a child process in a variable in the parent in NodeJS

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