'grep +A': print everything after a match
awk, bash, grep, sed
Solution
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
Problem
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.