Line 18 | 200 040 200
200 200 051 200 1C2 200 2E2 Line 18 |
200 040 200 040 200 052 200 1B9 200 2F4
Line 18 | 200 040 200
040 200 200 200 1C2 200 2DC
Line 18 | 200 040 200
040 200 063 200 1D6 200 2D4
How do I deleted per line all the 'odd' 200?
So that it becomes like:
Line 18 | 040
200 051 1C2 2E2 Line 18 |
040 040 052
1B9 2F4 Line 18 |
040 040 200 1C2 2DC
Line 18 | 040
040 063 1D6 2D4
use the 'g' flag of the substitute command:
:%s/\<200\>/ /g
Notice that his output does have *some* 200s in it. I'm not sure
what the input/output looks like exactly, as it's possible that
MTA/MUAs between Eric and the VimML and my machine have bunged
with matters. Getting the original text might help:
bash> head -3 file.txt | xxd | vi -
should give you a binary dump of the first 3 lines of file.txt
allowing us to better see where linebreaks and spaces/tabs fall.
It *sounds* like Eric is looking for something like this
monsterous one-liner:
%s/\<\(\x\x\x\)\(\_s\+\x\x\x\)\>/\=((submatch(1)==200)?'
':submatch(1)).submatch(2)
(that's 3 spaces in those quotes, in case of further mailer
problems, one for each character of "200")
which finds all the "odd" 200's in the block (as a ":hls" with
searching on that initial pattern will show it finds)
However, it doesn't come out with the same results as Eric's
expected results. Thus, we either have a problem of conveying
the problem, or there's an error in the expected output.
-tim
