How to grep for contents after pattern?

Question

Given a file, for example:

potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789

I\'d like to grep for all lines that

Answer 1

You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.

Try this (semicolons are there to allow you to put it all on one line):

$ while read line;
  do
    if [[ "${line%%:\ *}" == "potato" ]];
    then
      echo ${line##*:\ };
    fi;
  done< file.txt

## tells bash to delete the longest match of ": " in $line from the front.

$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789

or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.

$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi

The substring to split on is ":\ " because the space character must be escaped with the backslash.

You can find more like these at the linux documentation project.