how to get integer part from string in Shell Script

sunilvadranapu · 10-31-2008, 09:59 AM

Hi,
i wrote a script to get the product release. i am getting this correctly.
version=13q.00.00.00;
Now i need to get the major release which is 13 from 13m. How can i do this. please help me to get it work.

Thanks in advance.

-sunil

ErV · 10-31-2008, 10:14 AM

Quote:

Originally Posted by sunilvadranapu

How can i do this.

Use regular expressions with "grep" or "sed".

drchuck · 10-31-2008, 11:36 AM

or, using awk and bash:

Code:

major_version=$( $version |awk -F '.' '{print $1}' )

Do you want the answer to be 13, or 13q? You could strip off the final character, like this:

Code:

integer=${major_version:0:${#major_version}-1 }

colucix · 10-31-2008, 11:39 AM

You can try also parameter substitution in two passes:

Code:

version=13q.00.00.00
release=${version%%.*}
release=${release//[[:alpha:]]}
echo $release

The first substitution strips out the part of the string starting with a dot (.00.00.00), the second substitution removes all the alphabet characters.

rash31 · 11-02-2008, 04:51 AM

One way you can do this is by using the cut and expr commands. You can obtain the
string as 13m from your script output for eg say- "13m.00.00.00" by piping the output to cut command. Then, further you can store the output in the variable "version" through the script as version=13m and use the expr command as #expr "$version" : '$..$.' and the final output that you get is 13. I hope this will work for you.