bash script question

jbraner · 08-10-2007, 04:30 AM

Hi,

Please bear with me as this is all pretty new to me. I'm trying to write a (test) script where a user is prompted for input that must be a number. If it's not a number, I want to keep telling them to try again, until they get it right ;-)

For this reason, I tried to evaluate the input (called var1) in a while or until loop - but I have had no luck. The only way I can get regular expressions to work is by echoing var1 and then piping this to grep. Here is what I've got so far.

<snip>
echo "Type a whole number: "
read var1
# make sure we're getting a number as input
until test `echo $var1 | egrep -o [0-9]+`
do
echo "You have to type a number. Try again..."
read var1
done
# now strip out the non-numbers if they're there
var1=`echo $var1 | egrep -o [0-9]+`
</snip>
(I'm just taking this number and adding it to another number and printing the result on the screen - nothing earth shattering)

This part of the script will reject any input that does not contain any numbers, but it will still accept things like "55r" - so I used the "-o" switch for egrep to drop the "r" in this case.

What I would really like to do is reject *everything* that is not a number and make you "try again.." ;-)

Is there a regular expression for [0-9]+ but not letters (or other characters)? I tried things like [0-9^a-Z] but have had no luck.

Am I just getting the syntax wrong? Or do I have to nest a bunch of loops?
All the examples I can find for this sort of thing use "if .. then" statements, but this only works one time and will not keep making the user "try again"

I hope this makes sense.

John B

rupertwh · 08-10-2007, 04:47 AM

Like this?

Code:

#!/bin/bash

while true ; do
        echo "Enter a number: "
        read var
        if echo $var | egrep -q '^[0-9]+$' ; then
                break
        fi
        echo "'$var' is not a number. Try again..."
done
echo "Good"

jbraner · 08-10-2007, 05:37 AM

Rupert,

Thank you! That works perfectly!
If you don't mind - can i just ask one question?
egrep -q '^[0-9]+$'
this is returning the correct reply of only numbers
the regular expression reads (in English):
^[0-9] "the first character must be 0-9"
+ "matches one or more occurrrences after the initial one"
$ what does the $ do? I haven't seen this.

BTW I changed

if echo $var | egrep -q '^[0-9]+$' ; then

to

if `echo $var | egrep -q '^[0-9]+$'`
then
(added back tic and put then on seperate line)

Thanks again

jbraner · 08-10-2007, 05:53 AM

Quote:

Originally Posted by jbraner

$ what does the $ do? I haven't seen this.

sorry - I was being thick. The $ means that it must *end* with this characrter too.

So it must start with a number, have any number of numbers and end with a number.
I *knew* there was a way to do this - thanks again

pixellany · 08-10-2007, 07:34 AM

Quote:

Originally Posted by jbraner

sorry - I was being thick. The $ means that it must *end* with this characrter too.

So it must start with a number, have any number of numbers and end with a number.
I *knew* there was a way to do this - thanks again

In the interest of learning, let's split some hairs.....
"^" means "the empty space at the beginning of the line"
"$" means "the empty space at the end of the line"

SO:
"[0-9]$" means "any numeral, immediately followed by the empty space a the end of the line"

jbraner · 08-10-2007, 07:45 AM

pixellany,

All hair splitting is welcome. I have not seen it written this way before, but what you say makes perfect sense.

<still a *lot* more to learn>