'grep +A': print everything after a match

Question

I have a file that contains a list of URLs. It looks like below:

file1:

http://www.google.com
http://www.bing.com
http://www.yahoo.com
http://www.baidu.com
http://www.yandex.com
....

I want to get all the records after: http://www.yahoo.com, results looks like below:

file2:

http://www.baidu.com
http://www.yandex.com
....

I know that I could use grep to find the line number of where yahoo.com lies using

grep -n 'http://www.yahoo.com' file1

3 http://www.yahoo.com

But I don't know how to get the file after line number 3. Also, I know there is a flag in grep -A print the lines after your match. However, you need to specify how many lines you want after the match. I am wondering is there something to get around that issue. Like:

Pseudocode:

grep -n 'http://www.yahoo.com' -A all file1 > file2

I know we could use the line number I got and wc -l to get the number of lines after yahoo.com, however... it feels pretty lame.

Answer 1

AWK

If you don't mind using AWK:

awk '/yahoo/{y=1;next}y' data.txt

This script has two parts:

/yahoo/ { y = 1; next }
y

The first part states that if we encounter a line with yahoo, we set the variable y=1, and then skip that line (the next command will jump to the next line, thus skip any further processing on the current line). Without the next command, the line yahoo will be printed.

The second part is a short hand for:

y != 0 { print }

Which means, for each line, if variable y is non-zero, we print that line. In AWK, if you refer to a variable, that variable will be created and is either zero or empty string, depending on context. Before encounter yahoo, variable y is 0, so the script does not print anything. After encounter yahoo, y is 1, so every line after that will be printed.

Sed

Or, using sed, the following will delete everything up to and including the line with yahoo:

sed '1,/yahoo/d' data.txt

Answer 2

This is much easier done with sed than grep. sed can apply any of its one-letter commands to an inclusive range of lines; the general syntax for this is

START , STOP COMMAND

except without any spaces. START and STOP can each be a number (meaning "line number N", starting from 1); a dollar sign (meaning "the end of the file"), or a regexp enclosed in slashes, meaning "the first line that matches this regexp". (The exact rules are slightly more complicated; the GNU sed manual has more detail.)

So, you can do what you want like so:

sed -n -e '/http:\/\/www\.yahoo\.com/,$p' file1 > file2

The -n means "don't print anything unless specifically told to", and the -e directive means "from the first appearance of a line that matches the regexp /http:\/\/www\.yahoo\.com/ to the end of the file, print."

This will include the line with http://www.yahoo.com/ on it in the output. If you want everything after that point but not that line itself, the easiest way to do that is to invert the operation:

sed -e '1,/http:\/\/www\.yahoo\.com/d' file1 > file2

which means "for line 1 through the first line matching the regexp /http:\/\/www\.yahoo\.com/, delete the line" (and then, implicitly, print everything else; note that -n is not used this time).

Answer 3

awk '/yahoo/ ? c++ : c' file1

Or golfed

awk '/yahoo/?c++:c' file1

Result

http://www.baidu.com
http://www.yandex.com

Answer 4

This is most easily done in Perl:

perl -ne 'print unless 1 .. m(http://www\.yahoo\.com)' file

In other words, print all lines that aren’t between line 1 and the first occurrence of that pattern.

Answer 5

Using this script:

# Get index of the "yahoo" word
index=`grep -n "yahoo" filepath | cut -d':' -f1`

# Get the total number of lines in the file
totallines=`wc -l filepath | cut -d' ' -f1`

# Subtract totallines with index
result=`expr $total - $index`

# Gives the desired output
grep -A $result "yahoo" filepath