php exec() returning 1 when running node js command

Question

I need to run node js command from my php file. I am using exec() to do this. Here is my script

<?php

$ret = exec("node -v", $out, $err);
echo print_r($out);
?>

Here I am getting return value 1. How to fix this?

Is node in the PATH for the user/process running PHP? (E.g. run print_r(getenv('PATH')) to check / compare to path you get when running which node on terminal) — Tobias K.
– Tobias K., Commented Aug 10, 2018 at 12:04
$out contains ONLY STD_OUT, to see an err u shd run "/usr/bin/node foo.js 2>&1" — Vasilii Suricov
– Vasilii Suricov, Commented Jan 26, 2024 at 12:21

xate · Accepted Answer · 2018-08-10 12:12:55Z

1

Short summary:

$ret contains last line of result
$out array of result lines
$err "any non-zero return value is an error" (referencing to this one https://stackoverflow.com/a/2230620/3278963)

So 1 means "Miscellaneous errors", have you tried running the command in the shell to see what happens?

xate

6,3723 gold badges20 silver badges43 bronze badges

answered Aug 10, 2018 at 11:59

Cybeaer

439 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1