Help on extracting a substring from the input string

Shell Programming and Scripting
1

Hi,

I am new to Unix. I am trying to extract a substring from an input string: Ex -

input string: deploy_v11_9_1
i want to extract and store the value v11_9_1 from the input string in a new variable.

I am using following command in my shell script file:

echo "Enter the folder name u want to access ... ";
read FOLDER;
version = $(echo $FOLDER | awk '{split($0,a,"_v");print a[2]}');
echo $version;

However, I am getting error message that

version: command not found

Could you please help?

Thanks,
Pranav

2 
version = $(echo $FOLDER | awk '{split($0,a,"_v");print a[2]}');

Remove the spaces around the "=" sign:

version=$(echo $FOLDER | awk '{split($0,a,"_v");print a[2]}');
1 Like
3

We also need to remove '$' symbol and replace it with `.

Following code finally worked:

echo "Enter the folder name u want to access ... ";
read FOLDER;
version=`echo $FOLDER | awk '{split($0,a,"_v");print a[2]}'`
echo "Version is "$version

output:

Enter the folder name u want to access ...
deploy_v11_9_1
Version is 11_9_1
4 
version=`echo $FOLDER | awk '{split($0,a,"_v");print a[2]}'`

Here you also have removed the ; at the end compare to post above.

Try this and see if it works
version=$(echo $FOLDER | awk '{split($0,a,"_v");print a[2]}')
Its more easy to read a code using $(code) compare `code`

It may be shorten some to:

version=$(awk '{split($0,a,"_v");print a[2]}' <<< $FOLDER)

I do also not see why you split $0
This may also work

version=$(awk -F"_v" '{print $2}' <<< $FOLDER)
1 Like
5

As per your given input your final output is highlighted above.
If you want v in the final output then use sth like this..

And instead of using split just us FS for the same :slight_smile:

$echo "deploy_v11_9_1" | awk -F "y_" '{ print $2}' 
v11_9_1

or if you want only numbers then..

$echo "deploy_v11_9_1" | awk -F "_v" '{ print $2}' 
11_9_1

and please look this..

 
$ version=`echo "deploy_v11_9_1" | awk -F "_v" '{ print $2}'`
 
$ echo $version
11_9_1
 

$ version=$(echo "deploy_v11_9_1" | awk -F "_v" '{ print $2}')
 
$ echo $version
11_9_1

Both $() and `` are used for the same but $() is preferred over backticks.

1 Like
6

Thanks all for ur valuable suggestions :slight_smile: