Input separator of the `read` builtin in Bash

Question

Given such situation:

echo "Please enter your name"
read name    # user enters: john smith
echo $name   # prints: john

What could cause read to read only the first word of the input? Is there a shell variable that controls this?

I came across this in a question on Ask Ubuntu and I'm wondering how to reproduce this behavior.

It has to have something to do with multiple fields - $IFS or filename generation. But echo concats on spaces - and so even for multiple args they should come together. I can imagine if it was an extra quote level deep - like an ssh script or something it might lose some - but that's fuzzy. echo should at least print the args unless $IFS is composed of all of the alphabetics or something. printf, on the other hand, would only print john or 123 in that scenario, though. — mikeserv, Dec 27, 2014 at 23:17

glenn jackman · Accepted Answer · 2014-12-28 00:32:07Z

1

to accept whatever the user enters, use this form

IFS= read -r name

That will accept leading/trailing/inner spaces as well as literal backslashes.

answered Dec 28, 2014 at 0:32

glenn jackman

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