Pattern Search between specific lines and print line numbers

grepsed

I have a file of 20 lines. I want to search for a string between the lines 10 and 15 and print the string along with the line numbers of the original file.

I have tried sed -n 10,15p file | grep -n "pattern"

The problem with the above command is, the output line number 10 of sed will be line number 1 for grep.

Hence it is not printing the line numbers from the original file.

Best Answer

When working w/ sed I typically find it easiest to consistently narrow my possible outcome. This is why I sometimes lean on the !negation operator. It is very often more simple to prune uninteresting input away than it is to pointedly select the interesting kind - at least, this is my opinion on the matter.

I find this method more inline with sed's default behavior - which is to auto-print pattern-space at script's end. For simple things such as this it can also more easily result in a robust script - a script that does not depend on certain implementations' syntax extensions in order to operate (as is commonly seen with sed {functions}).

This is why I recommended you do:

sed '10,15!d;/pattern/!d;=' <input

...which first prunes any line not within the range of lines 10 & 15, and from among those that remain prunes any line which does not match pattern. If you find you'd rather have the line number sed prints on the same line as its matched line, I would probably look to paste in that case. Maybe...

sed '10,15!d;/pattern/!d;=' <input |
paste -sd:\\n -

...which will just alternate replacing input \newlines with either a : character or another \newline.

For example:

seq 150 |
sed '10,50!d;/5/!d;=' |
paste -sd:\\n -

...prints...

Related Solutions

Search a pattern and print preceding lines starting with another pattern

Here's a solution in Perl:

perl -nlE '
    if    (/a/)   { @buffer = ($_) }
    elsif (/xyz/) { push @buffer,$_; say for @buffer }
    else          { push @buffer,$_}
' your_file

How this works

It reads through the file line-by-line and does one of three things:

If the current line matches the pattern a, it assigns the current line to the @buffer array.
If the current line matches the pattern xyz, it pushes the current line onto the buffer and prints the contents of the buffer
If none of the two cases above is true, it simply appends the current line to the @buffer array.

Thus, whenever a new line matches the pattern a, the contents of the @buffer are erased and replaced by the current line only. This guarantees you will find the closest a preceding xyz.

You should of course replace the regexes I used with the actual regexes relevant to your case.

Grep with Line Numbers – Output Neighboring Lines

You can use GNU grep's Context Line Control, from man grep:

-A NUM, --after-context=NUM
Print NUM lines of trailing context after matching lines. Places a line containing a group separator (--) between contiguous groups of matches. With the -o or --only-matching option, this has no effect and a warning is given.

-B NUM, --before-context=NUM
Print NUM lines of leading context before matching lines. Places a line containing a group separator (--) between contiguous groups of matches. With the -o or --only-matching option, this has no effect and a warning is given.

If you require an equal number of lines either side, you can use a single context number:

-C NUM, -NUM, --context=NUM
Print NUM lines of output context. Places a line containing a group separator (--) between contiguous groups of matches. With the -o or --only-matching option, this has no effect and a warning is given.

When combined with the -n flag (--line-number), you have numbered matches with surrounding context.

In your example:

grep -n -C1 download

Best Answer

Related Solutions

Search a pattern and print preceding lines starting with another pattern

Grep with Line Numbers – Output Neighboring Lines

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