UNIX for Dummies Questions & Answers

Hello, i have a script which checks if the user entered 8 numeric characters in the form of YYYYMMDD (birth date). If the user entered any non numeric characters, an error will be displayed:

Code:

# Check to see if the 8 characters are all numbers
# If not show error essage
# And prompt user for more input

         echo $char | grep -q '^[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]$'  
if 	[ $? -ne 0 ]
then
         echo "You have entered non-numeric values.  Please type in the form of YYYYMMDD"
	read char
        continue

Is there a simpler way to write this command without using the [0-9] value for each field? Any help would be appreciated

On Solaris you can use:

Code:

echo "12345678" | /usr/xpg4/bin/grep -Eq "^[0-9]{8}$"

In any case, you must use a tool which supports extended regular expressions to use the above regex.

Hi,

What if the numbers are comma separated and the no.of occurences of the pattern is not known.
E.g. 1,2,3,4,...n

TIA

@12345 If i understood your question correctly, then the regex is:

Code:

grep -E "^[0-9,]{3,}"

Hi,

Thanks for your reply. I just want to make this requirement more clear.
I read values to a variable, using "read var".I am now validating the user input. The user should input the values in the format 1,3,6 (can enter any numeric values upto n).
I have tried this for 2 numbers. This is the command that I had used.
read col
1,2 ----> input provided by user
echo $col | grep -w "^[0-9],[0-9]$"
return code is successful.
If it contains any other input like 1,a or a,1 or a,b it returns a code of 1.

This code works fine for 2 numbers, but I want to make this work for 'n' numbers.
I had tried the solution provided by you, but -E command is not recognised by my script.
TIA