How to print a range of character in unix using cut and grep?

Question

I can print block of characters using (cut -d) specifying delimiter.

# grep -i access.log| grep ' 500 '| cut -d\- -f1 

But how can I print characters after specific pattern "address=tel%3A%2B"

For example: how to print the entire phone numbers in this file?

begining of the line address=tel%3A%2B416xxxxxxx rest of the log
begining of the line address=tel%3A%2B647xxxxxxx rest of the log
begining of the line address=tel%3A%2B519xxxxxxx rest of the log

...

Answer 1

sed -e 's,.*address=tel%3A%2B\([^ ]\+\).*,\1,g' <<EOF                          
begining of the line address=tel%3A%2B416xxxxxxx rest of the log                
begining of the line address=tel%3A%2B647xxxxxxx rest of the log                
begining of the line address=tel%3A%2B519xxxxxxx rest of the log                
EOF

Output:

416xxxxxxx
647xxxxxxx
519xxxxxxx