sed + cut words before the first “END” word in line

Question

Hi all

How to cut words before the first “END” word in line, and not include the first character before the “END” word ( in our examples the character is “-“)

Only with sed command

For example:

  echo AAA-BBB-CCC-END | sed ...
  AAA-BBB-CCC


  echo 1-END-3-4 | sed ...
  1     


  echo 1-2-END-3-4 | sed ...
  1-2


  echo PARAM1-PARAM2-END | sed ...
  PARAM1-PARAM2


  echo fileExample1-fileExample2-END | sed ...
  fileExample1-fileExample2


  echo xyz_123-END | sed ...
  xyz_123

Answer 1

sed "s/.END.*//"

This should do the trick.

It'll remove the first char before the END, and all chars after (and of course the END itself).

Answer 2

sed -r "s/(.*).END.*/\1/"

would be my suggestion

Answer 3

if your string structure always have hyphens and then the "END" word (eg -END),

$ var="AAA-BBB-CCC-SEND-END"
$ echo ${var%%-END*}
AAA-BBB-CCC-SEND

As you can see, in case you have words like "SEND", it will still give you the correct output.

if you still prefer using sed,

$ echo $var | sed 's/-END.*//'
AAA-BBB-CCC-SEND

Answer 4

this seems to work: awkish not sedish, & literal

awk '$0 !~ /(echo*|^$)/'