My answer is based on previous answers, but with the addition of reversing the sequence so that history items are deleted from most recent to least recent.
Get your current history (adjust the number of lines you want to see):
history | tail -n 10
This gives me something like
1003 25-04-2016 17:54:52 echo "Command 1"1004 25-04-2016 17:54:54 echo "Command 2"1005 25-04-2016 17:54:57 echo "Command 3"1006 25-04-2016 17:54:59 echo "Command 4"1007 25-04-2016 17:55:01 echo "Command 5"1008 25-04-2016 17:55:03 echo "Command 6"1009 25-04-2016 17:55:07 echo "Command 7"1010 25-04-2016 17:55:09 echo "Command 8"1011 25-04-2016 17:55:11 echo "Command 9"1012 25-04-2016 17:55:14 echo "Command 10"
Select the start and end positions for the items you want to delete. I'm going to delete entries 1006 to 1008.
for h in $(seq 1006 1008); do history -d 1006; done
This will generate history -d
commands for 1006, then 1007 becomes 1006 and 1006 is deleted, then 1008 (became 1007) is now 1006 and gets deleted.
If I also wanted to delete the history delete command then it's a bit more complicated because you need to know the current max history entry.
You can get this with (there may be a better way):
history 1 | awk '{print $1}'
Putting it together you can use this to delete a range, and also delete the history delete command:
for h in $(seq 1006 1008); do history -d 1006; done; history -d $(history 1 | awk '{print $1}')
Wrap this all up in a function to add to your ~/.bashrc
:
histdel(){ for h in $(seq $1 $2); do history -d $1 done history -d $(history 1 | awk '{print $1}') }
Example deleting command 4, 5 and 6 (1049-1051) and hiding the evidence:
[18:21:02 jonathag@gb-slo-svb-0221 ~]$ history 11 1046 25-04-2016 18:20:47 echo "Command 1" 1047 25-04-2016 18:20:48 echo "Command 2" 1048 25-04-2016 18:20:50 echo "Command 3" 1049 25-04-2016 18:20:51 echo "Command 4" 1050 25-04-2016 18:20:53 echo "Command 5" 1051 25-04-2016 18:20:54 echo "Command 6" 1052 25-04-2016 18:20:56 echo "Command 7" 1053 25-04-2016 18:20:57 echo "Command 8" 1054 25-04-2016 18:21:00 echo "Command 9" 1055 25-04-2016 18:21:02 echo "Command 10" 1056 25-04-2016 18:21:07 history 11[18:21:07 jonathag@gb-slo-svb-0221 ~]$ histdel 1049 1051[18:21:23 jonathag@gb-slo-svb-0221 ~]$ history 8 1046 25-04-2016 18:20:47 echo "Command 1" 1047 25-04-2016 18:20:48 echo "Command 2" 1048 25-04-2016 18:20:50 echo "Command 3" 1049 25-04-2016 18:20:56 echo "Command 7" 1050 25-04-2016 18:20:57 echo "Command 8" 1051 25-04-2016 18:21:00 echo "Command 9" 1052 25-04-2016 18:21:02 echo "Command 10" 1053 25-04-2016 18:21:07 history 11
The question was actually to delete the last 10 commands from history, so if you want to save a little effort you could use another function to call the histdel
function which does the calculations for you.
histdeln(){ # Get the current history number n=$(history 1 | awk '{print $1}') # Call histdel with the appropriate range histdel $(( $n - $1 )) $(( $n - 1 )) }
This function takes 1 argument, the number of previous history items to delete. So to delete the last 10 commands from history just use histdeln 10
.