Unix grep/test command

netmaster · #1 (**permalink**) 10-08-2007

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

robotronic · #2 (**permalink**) 10-09-2007

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.

12345 · #3 (**permalink**) 05-07-2009

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

thegeek · #4 (**permalink**) 05-07-2009

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

Code:

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

12345 · #5 (**permalink**) 05-08-2009

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

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