[SOLVED] Bash Regex extract all text from 2nd occurence of specific character until end of line

Issue

I have the following strings:

text/:some_random_text:text_i_w4nt_to:k33p.until_th3_end_1
text/:some_random_text:text_i_w4nt_to::k33p.until_th3_end_1

with a regular expression, I want to extract:

text_i_w4nt_to:k33p.until_th3_end_1
text_i_w4nt_to::k33p.until_th3_end_1

I have tried with regex101.com the following expression: ([^:]+)(?::[^:]+){1}$ and it worked (only for the first string)

But if I try in bash, it does not

echo "text/:some_random_text:text_i_w4nt_to::k33p.until_th3_end_1" | sed -n "/([^:]+)(?::[^:]+){1}$/p"

Solution

Using sed

$ sed s'|\([^:]*:\)\{2\}\(.*\)$|\2|' input_file
text_i_w4nt_to:k33p.until_th3_end_1
text_i_w4nt_to::k33p.until_th3_end_1

or

$ sed s'|\([^:]*:\)\{2\}||' input_file
text_i_w4nt_to:k33p.until_th3_end_1
text_i_w4nt_to::k33p.until_th3_end_1

Answered By - HatLess

Answer Checked By - Timothy Miller (WPSolving Admin)