Shell Programming and Scripting

This might sound easy but its not...

I have a file a with the following data...including the 1,2 and 3

1 this.is.it
2 is.it
3 this.is

now i need to grep for "is.it" ONLY in the file without any data manipulation like grep -v,cut etc,. In other words my output should be "is.it" only

How do i do it?

You have 28 posts and you don't know how to use code tags or how to explain your problem ?
Try again please.

If your grep has the -o option, use that. A brief reading of the manual page should have told you that already.

Code:

grep -o 'is\.it' file

The backslash is necessary because the dot has a special meaning in regular expressions (it matches any one character, except newline).

Hi Era,

Thanks for your reply...but with grep -o 'is\.it' file , the output would be as follows

is.it
is.it

that is "is.it" appears twice in the output...I know we can do an uniq to get it only once but as I said earlier I dont want to do any data manipulations. Is there any direct command which can give the desired output?

It's not clear on what criteria it should be unique. Do you mean grep lines matching only that text? There is no way in grep to make the output unique; the "data manipulations" as you call them are inherent to the design philosophy of these tools.

This searches for "is.it" and no other text:

Code:

grep '^[^A-Za-z]*is\.it[^A-Za-z]*$' file

The expression [^A-Za-z]* means any number of (that's the trailing asterisk; "any number" includes zero) characters which are not A through Z or a through z. Caret (^) matches beginning of line, and dollar sign ($) matches end of line. Hence, any line containing "is.it" and no other alphabetics before or after.

However, with -o the whole match will be printed, and in that case, this means the whole line. You can use sed to make substitutions on the same pattern:

Code:

sed -n 's/^[^A-Za-z]*\(is\.it\)[^A-Za-z]*$/\1/p' file

This replaces the whole line with just the part between the \( parentheses \) and then prints the resulting line.

Tell us again why you need this to be a single command?

Try awk:

Code:

awk '$2=="is.it"{print $2}' file

Thanks Era and Danmero..

Both your solutions work great! but i would go for the grep option as i am using the grep -f option which is not available with awk.

Era.. your grep solution can be further shortened to this..

grep '^[^A-Za-z]*is\.it$' file

Thank you!