I'd like to replace all line containining "CreateTime=xxxxx" with "CreateTime=2012-01-04 00:00". May I know how should I do it with vim?
[m18]
Attendees=38230,92242,97553
Duration=2
CreateTime=2012-01-09 22:00
[m20]
Attendees=52000,50521,34025
Duration=2
CreateTime=2012-01-09 00:00
[m22]
Attendees=95892,23689
Duration=2
CreateTime=2012-01-08 17:00
You can use the global substitute operator for this.
:%s/CreateTime=.*$/CreateTime=2012-01-04 00:00/g
You can read the help for the s command from within Vim using:
:help :s
You can read about patterns with :help pattern-overview.
As requested, a bit more about the regular expression match (CreateTime=.*$):
CreateTime= # this part is just a string
. # "." matches any character
* # "*" modifies the "." to mean "0 or more" of any character
$ # "$" means "end of line"
Taken together, it matches CreateTime= followed by any series of characters, consuming the rest of the line.
Related
Sometimes I need to edit files which should not end with a newline.
However vim\nvim by default do not visualise in any way the newline character at the end of file. Therefore I am not able to:
visually confirm if the file has a newline character at the end or not
remove that character
Are there any setting which would allow me to see the tailing newline character and edit it in the same way as any other characters?
For example, after create 2 files as follows:
echo test > file-with-newline
echo -n test > file-without-newline
opening first one with nvim file-with-newline shows:
test
~
~
file-with-newline
opening second one with nvim file-without-newline shows:
test
~
~
file-without-newline
Navigating with the cursor to the end of line in either case yields the same result (the cursor stops after last visible character: t). There is no way to tell if the newline is there or not, let alone remove it using familiar commands used to remove ordinary characters (or newlines within the file).
You can enable the option :help 'list':
:set list
to show that "newline character" as a $ at the end of the line (among other things):
Note, however, that the option doesn't make the character "editable" in any way.
if the file has a newline character at the end or not
:set eol?
endofline
remove that character
:set noeol nofixeol
:update
how do i do this in vi?
awk -F"," awk '{print substr($1,1,10)}'
I only want to keep the first 10 characters of my date column (example 2014-01-01) and not include the timestamp.
I tried to do it in awk but i got this error:
sed: RE error: illegal byte sequence
I believe it is a bash_profile setting error.
This is what i have in my bash_profile:
#export LANG=en_US.UTF-8
#export LOCALE=UTF-8
export LC_CTYPE=C
export LANG=C
in vim, do:
:%norm! 11|D
this will affect all lines in your buffer.
If you like, :s could do this job too.
:%s/.\{,10}\zs.*//
:%s/: apply the substitution to all the lines
.\{,10}: match anything up to 10 times (greedy)
\zs: indicates the beginning of the match
.*: match the rest of the line
/: end of the first part of :s
/: end of the second part of s (since there's nothing between the two /, replace with nothing, ie delete)
For editing blocks of text there is a -- VISUAL BLOCK -- mode accessible via CTRL-V (on Windows ussually CTRL-Q). Then you can press d to delete your selection.
Or with a simple substitute command
:%s/\%>10c.*//
\%>10c - matches after tenth column
. - matches any single character but not an end-of-line
* - matches 0 or more of the preceding atom, as many as possible
Or you can use range
:1,3s/\%>10c.*//
This would substitute for the first three lines.
What is the special character which indicate first ?
if we do
$ vi .bashrc
$ source !$
this !$ will replaced by .bashrc
because ! means previous line(am I correct?), $ means last word (for sure)
then what is first?
I want to insert some string in every line in vi editor using
:%s/find-key-word/replaced-keyword/g
in here, if I put
:%s/$/example/g
in vi editor, it will append in all lines with example.
I want to insert all in front of all string every line.
I know I can use visual block (ctrl+v) and select all front lines and insert (shift+i) insert some word and escape(esc) will do the same... but I want to do in one shot..
please let me know how to do..
Thanks in advance
There are two questions, so you are getting two kinds of answers :)
The bash command history has only a passing similarity to the vi regular expression syntax.
^ is the beginning of line in vi. $ is the end of line in vi.
!!:0 is one way of accessing the first word of the previous command in bash
!$ is one way of accessing the last word of the previous command in bash
To indicate beginning of line, the symbol used is:
^
See an example:
$ cat a
hello!
this is me
testing some
stuff
$ sed 's/^/XXX/' a
XXXhello!
XXXthis is me
XXXtesting some
XXXstuff
The character you are looking for is ^.
For example, :%s/^/example/g will prepend all lines with the string example.
In bash, !^ refers to the first argument of the previous command, and !$ the last argument.
There is a text file like that
root:x:0:0:root:/root:/bin/bash
bin:x:1:1:bin:/bin:/bin/false
daemon:x:2:2:daemon:/sbin:/bin/false
mail:x:8:12:mail:/var/spool/mail:/bin/false
ftp:x:14:11:ftp:/srv/ftp:/bin/false
http:x:33:33:http:/srv/http:/bin/false
How to replace in VI all characters in strings 2-4 from 2nd to 5th to 'X'?
UPD:
it's smth like: :2,4s//X/g
I guess I need a regular expression
UPD2: :2,4s/^\(.\)...\|^$/\1XXX/ | 2,4s/^$/ XXX/
Vim only
Try this command:
:2,4s/\%2c.../XXX/
Where:
2,4 is for in strings 2-4
\%2c... is for from 2nd to 5th
XXX is for to 'X'
Both Vim and Vi
As Vi doesn't have \%c, this command should be used instead:
:2,4s/^\(.\).../\1XXX/
Resources
:help :range
:help /ordinary-atom
I have a text file that's thousands of lines long. Every line starts with a string of 8 hex numbers. I need to remove this string on every line. How do I do this in vim?
Use ^V for block select, highlight your eight columns, and delete as normal.
Or use :s:
:%s/\v^[a-fA-F0-9]{8}//
Replace first 8 hex chars (0-9 digits, a-f/A-F letters) on any line with empty string:
:%s/^[0-9a-fA-F]\{8\}//gc
If the line is
12345678 Something else
a total of 9 chars is to be removed from the head of each line, in VIM
:1,$s/^.........//
should do the trick (9 dots),
: to tell vim you want to enter a command
1,$ means the command affects from line 1 to the last (or g global)
s means substitute
^ means beginning of line
..... means 5 (any) chars
s/^.....// means replace 5 chars at start of line with nothing
edit to match the number of hex chars from the question..
use cut command. This way is much more straight forward.
echo '12345678 Something else' | cut -c 10-
result:
Something else
Just remind, cut index string start from 1 instead of 0.
In vi, we could just run cut in vi:
:%!cut -c 10-