CODEWITHSUNDEEP
mongodb
sudesh
sudesh

Reputation: 973

Skip printing to stdout when using "mongo --eval"

I need the js variable to be passed through command line arguments. The option --eval did this. But it also printing the variable value to the shell by default. I want to skip the variable value printed to the shell.

Ex:-
I have used "user_name" inside the script.js

Passing the value using mongo --eval="user_name='john'" script.js
But it un necessarily printing the value
  john

Any ideas.

Upvotes: 0

Views: 919

Answers (2)

Tomas
Tomas

Reputation: 5143

For this use case you could use var in your declaration. This will not print the value:

mongo --eval="var user_name='john';" script.js

Upvotes: 1

Ross
Ross

Reputation: 18111

There is a --quiet flag - but evaluated items are still printed to stdout.

You could pipe to /dev/null or a file to capture the stdout eg:

mongo --eval="user_name='john'" script.js > /dev/null

Upvotes: 1

Related Questions