Find and replace whole words in vim - regex

To find and replace all instances of a word in vim, I use
%s/word/newword/g
How do I change this so that it only finds instances of "word" that are whole words?


You can use \< to match the beginning of a word and \> to match the end:
%s/\<word\>/newword/g


For case-sensitive replace.. you can use "\C"
:%s/\<word\>\C/newword/g
It replaces only "word" with newword leaving others like Word,WORD... unreplaced.


For PCRE compatible search and replace, you can use the perldo or rubydo commands as described here: http://vim.wikia.com/wiki/Perl_compatible_regular_expressions
For example:
:perldo s/\bword\b/newword/g

Related

Regular expression Notepad++

Good Morning in my timezone
I want to replace a character that is in the beginning of each line
So i had used the following regular expression to find the text
^\d
And it works fine in finding all the characters
The problem is in the replace with
I want to replace with single quote followed by the same character found above
How can i do it ?
Thanks in advance

You may try this option:
Find:
^(?=\d)
Replace:
' <-- just a single quote
The find pattern uses a positive lookahead which asserts that the first character is a digit, but nothing is ever matched. Then, the replacement is a single quote.

You may use
Find What: ^\d
Replace With: '$0
where $0 is the backreference to the match value.

Another one would be:
Find:
^(\d)
Replace:
'\1
In this example \1 would be 1st captured group.

How in notepad++ find/replace text between slashes?

How to find the text between the second and fourth slashes in a path like /folder/subfolder-1/subfolder-2/subfolder-3? I’m trying to replace this with something like /folder/new-folder/subfolder-3.
The most important for me is to be able to find the part after the n-th slash.
I tried the regex /((.*?)/){3}, but it doesn’t work.

Using Match resetter \K meta-character you are able to do it in a simpler way.
Find:
/.*?/\K(.*?/){2}
Replace with:
new-folder/

One way you could to it is by using this string in the pattern to replace
(/.+?)(/.+?){2}(/\S+)
And use this one in your pattern to replace it with
$1/new-folder$3
From your string:
/folder/subfolder-1/subfolder-2/subfolder-3
(/.+?) will match /folder as $1
(/.+?){2} will match /subfolder-1/subfolder-2 as $2 (not used)
(/\S+) will match everything that isn't a space, in this case/subfolder-3 as $3
Leaving you room to insert your new-folder in-between.

How can I just mark till the slash?
Find what: (/[^/]+/)[^/]+/[^/]+
Replace with: $1new-folder

To find text between second and forth slash you can use the regex ^(/[^/]*/)([^/]*/[^/]*) then you can reference to the text between slashes with \2 when replacing the text.
To keep the text before the slashes you can enter something like \1myNewTextBetweenSlashes2and4.

In notepad++ Find by this:
(/[^/]+)(?:/[^/]+/[^/]+/)(.*)
And Replace by this:
\1\/new-folder/\2
Make sure that: .matches newline is not checked

{2} indicates 2 levels after first level will be repalced by new-folder
Find:
(\/.*?\/)(.*?\/){2}(.*)
Replace:
$1new-folder/$3
Demo: https://regex101.com/r/XIA3IN/3

Notepad++ Regex - Find multiple characters and replace them with one character

I have a file with +20K lines and some strings have this structure:
,"/d/s/aaa.jpg","/e/_/bbb.jpg","/_/2/bbb.jpg" ....
and I want to replace them with:
,"/aaa.jpg", "/bbb.jpg","/bbb.jpg" ...
Can some one provide me a regex expression that will find those 5 leading characters and replace them with "/"?
Thank You in advance.

Use the following:
Find what: /[^/]+/[^/]+(/[^/]+\.jpg)
Replace with: $1

Edited:
The following:
\"\/[^/]+\/[^/]+\/
will match the "/d/s/ or "/e/_/ part of a string. You can test it here: http://regexpal.com/
Make sure to replace it with:
"/
to do more you would need capture groups (i.e. capturing some parts of the regex to reuse that in the substitution or manipulate it somehow)

How to find occurences of same subsequent characters in a string with a regular expression?

How can I find occurences of same subsequent characters in a string with a regular expression or function?
Example:
I am l​ee​t and I have a thr​ee​ pi​zz​as. That n​oo​b right there has only one pi​zz​a. P​oo​r boy.

You can use a backreference:
/(.)\1/
Change \1 to \1+ if you want to find sequences of length two or more.
Note that the syntax can vary depending on the regular expression engine you are using.

Not sure which version of regex you're working with, but for egrep, this works:
egrep '(.)\1' < file
That will show all lines that have two of some character in a row. If you want just letters:
egrep `([A-Za-z])\1' < file
would work.

Like this in a perl flavour. \w matches a word character, and \2 matches second parentheses.
m/((\w)\2+)/g

Google it:'double characters regex'
Here's a re-fiddle I made with your regex: http://refiddle.com/2fa

This should work ................ (.)\1+

vim regex find and replace

I want to use a regex to replace some strings in my file. I search for:
%s/^ [a-z]*/ /
what I want to do is to replace every [a-z]* that have 2 whitespaces with the sane [a-z] prepended with 4 whitespaces. Is there any "inplace" replacement or how would I reach that with vim?
With best regards

:%s/ \([a-z]*\)/ \1/g
should do the job; beware of running this multiple times, though because the result of the replace will match the input pattern :)

I find it more straightforward to use the \ze object to define the end of the match:
:%s/ \ze[a-z]*/ /g
so the [a-z]* is not included in the replace, but just used to match the relevant spaces.