Rund NodeJS file in shellscript loop

Question

My shellscript looks like

for i in {1..5}
    do
        echo "Welcome $i times"
        exec node nodefile.js
done

and the nodefile.js looks like

console.log 'running nodefile'

When I call the ./shellscript.sh there will only be displayed "Welcome x times" and "running nodefile" once.

Whats missing there? Is there anything that I have to return in nodefile or something else?

Mritunjay · Accepted Answer · 2014-07-23 07:33:06Z

1

Try this

for i in {1..5}
do
   echo "Welcome $i times"
   node process.js
done

answered Jul 23, 2014 at 7:33

Mritunjay

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1 Comment

Thanks a lot. Sometimes it could be so easy :)

TColbert · Accepted Answer · 2014-07-23 15:30:27Z

0

As an alternative you might try using the 'sha-bang' method.

Put this in your nodefile.js:

#!/the/location/of/your/node
console.log("running nodefile");

Then make the nodefile.js executable by doing:

chmod +x nodefile.js

Now your shell script can look something like this:

for i in {1..5}
do
   echo "Welcome $i times"
   /location/of/nodefile.js
done

answered Jul 23, 2014 at 15:30

TColbert

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