'grep +A': print everything after a match [duplicate]

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