Timeline for awk: print last N columns but with a different field separator
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 11, 2020 at 6:38 | answer | added | Kusalananda♦ | timeline score: 3 | |
| Apr 11, 2020 at 1:35 | answer | added | Ed Morton | timeline score: 2 | |
| Apr 10, 2020 at 22:19 | history | became hot network question | |||
| Apr 10, 2020 at 16:02 | comment | added | Santhosh | changed to seven. its not six | |
| Apr 10, 2020 at 16:00 | history | edited | Santhosh | CC BY-SA 4.0 | added 495 characters in body |
| Apr 10, 2020 at 15:20 | review | Close votes | |||
| Apr 15, 2020 at 3:05 | |||||
| Apr 10, 2020 at 15:07 | answer | added | αғsнιη | timeline score: 2 | |
| Apr 10, 2020 at 15:03 | comment | added | user313992 | Please make a little effort to tidy up your question. Do you really want the "last six columns" as in the text, or the last 7, as in the first example? Or the last NF-5 = 8 (13-5=8), as in the second example? Or was it that the last 5 (10-5), since you have changed the field definition)? | |
| Apr 10, 2020 at 14:45 | answer | added | RudiC | timeline score: 3 | |
| Apr 10, 2020 at 14:41 | comment | added | Kusalananda♦ | You two outputs shows that you either want the - in the date replaced by ::, or you won't. Which is it? Also, 2020::04::10::13::40::27::395533885 looks like seven columns, not six, and 6::2020-04-10-13::40::27::395533885 is five. | |
| Apr 10, 2020 at 14:33 | history | edited | terdon♦ | CC BY-SA 4.0 | edited tags; copy edit |
| Apr 10, 2020 at 14:33 | answer | added | terdon♦ | timeline score: 6 | |
| Apr 10, 2020 at 14:28 | answer | added | user000001 | timeline score: 2 | |
| Apr 10, 2020 at 14:11 | history | asked | Santhosh | CC BY-SA 4.0 |