Timeline for In bash, how can I echo the variable name, not the variable value? [closed]
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 25, 2022 at 8:05 | comment | added | lonix |
Shouldn't have been closed. He's asking for the equivalent of nameof in various programming languages. There are uses for that.
|
|
| Jul 12, 2018 at 10:57 | review | Reopen votes | |||
| Jul 12, 2018 at 13:53 | |||||
| May 3, 2018 at 23:21 | comment | added | user6860 | Related: unix.stackexchange.com/questions/397586/… | |
| May 23, 2017 at 12:40 | history | edited | CommunityBot |
replaced http://stackoverflow.com/ with https://stackoverflow.com/
|
|
| Dec 12, 2016 at 18:23 | comment | added | bubakazouba | I'm really curious, please tell us what did you need that for! | |
| May 13, 2014 at 5:13 | history | closed |
phemmer slm♦ Gilles 'SO- stop being evil' Anthon Thomas Nyman |
Needs details or clarity | |
| May 12, 2014 at 21:04 | answer | added | mjturner | timeline score: 37 | |
| May 12, 2014 at 21:04 | vote | accept | Andrew | ||
| May 12, 2014 at 20:47 | answer | added | xae | timeline score: 38 | |
| May 12, 2014 at 19:50 | comment | added | Renan Vicente | I don't get it too but it's what are you trying to do? echo '$var' | |
| May 12, 2014 at 19:41 | review | Close votes | |||
| May 13, 2014 at 5:13 | |||||
| May 12, 2014 at 18:22 | comment | added | iruvar |
I think you've missed the drift of the SO solution; try var=Test; Test=SO; echo ${!var}
|
|
| May 12, 2014 at 18:15 | comment | added | terdon♦ | I don't really understand the point of this. As @Braiam points out, since you are defining the name of the variable in the first place you will always know it and always be able to print it. What are you actually trying to do? | |
| May 12, 2014 at 18:14 | comment | added | Braiam |
Err... just echo var?
|
|
| May 12, 2014 at 18:14 | history | edited | Braiam |
edited tags
|
|
| May 12, 2014 at 18:10 | history | asked | Andrew | CC BY-SA 3.0 |