I have some text containing . A....
The regex is [.][\s][\s][!A-Z]
This finds the containing string. The problem is, that I don't know how to replace all of it with \n except the [A-Z]
You can use grouping and back refer to them by \1, \2, etc.
Change your regex to [.][\s][\s]([!A-Z]) and replace it with \n\1
I have a document and I want to replace all the paragraph return with the space.
I tried to replace it with space but unfortunately it did wrong like in the attache image it found space and A in search and it replace it with space nd.
What happened is that heaven." "
And
is replaced by heaven. nd
But I want it to be
heaven." "And
In MS Word Search and Replace, you may use groups and refer to their captured contents via backreferences like \1, \2, etc.
So, if you want to match a paragraph break that is followed with an ASCII letter, you need to use
^013([A-Za-z])
and replace with \1 (a space + \1, where \1 inserts the letter).
To find and replace * you can just use a normal replacing mode (uncheck Wildcard option). If you need to use * as a literal symbol in the wildcard pattern, put it into a []: [*].
I would like to change the second forward slash, within each line, to a comma.
I have found various posts and managed to derive a way of doing it from them but it's not doing it how I want.
Initial attempt - I thought I needed to replace between 2 delimiters
1st "Replace 2nd occurrence" - Found this post which seemed easier.
2nd "Replace 2nd occurrence"- Used the regex in here as a base for mine.
What I am doing is;
Find:
^(.*?)\/(.*?)\/
Replace:
$&,
Which results in changing my data from;
042146/OVERNIGHT/HSSC825571,started,14/07/2016,00:00:56,V0700LWHSB
042146/OVERNIGHT/HSSC825571,ended,14/07/2016,00:00:56,
042147/OVERNIGHT/HSSC825571,started,14/07/2016,00:00:58,V0700LWHSB
042147/OVERNIGHT/HSSC825571,ended,14/07/2016,00:00:58,
To;
042146/OVERNIGHT/,HSSC825571,started,14/07/2016,00:00:56,V0700LWHSB
042146/OVERNIGHT/,HSSC825571,ended,14/07/2016,00:00:56,
042147/OVERNIGHT/,HSSC825571,started,14/07/2016,00:00:58,V0700LWHSB
042147/OVERNIGHT/,HSSC825571,ended,14/07/2016,00:00:58,
Is there a way of just replacing the second /?
An example set of my data is;
042146/OVERNIGHT/HSSC825571,started,14/07/2016,00:00:56,V0700LWHSB
042146/OVERNIGHT/HSSC825571,ended,14/07/2016,00:00:56,
042147/OVERNIGHT/HSSC825571,started,14/07/2016,00:00:58,V0700LWHSB
042147/OVERNIGHT/HSSC825571,ended,14/07/2016,00:00:58,
042154/TEMP56/QPADEV000M,started,14/07/2016,00:01:02,V0700LRFIN
042154/TEMP56/QPADEV000M,ended,14/07/2016,00:07:12,
042155/JMALICKA/QPADEV000N,started,14/07/2016,00:01:05,V0700LRFIN
042155/JMALICKA/QPADEV000N,ended,14/07/2016,00:06:53,
042156/DG8SVCPRF/DG8SVC,started,14/07/2016,00:01:15,DATAGATE
042156/DG8SVCPRF/DG8SVC,ended,14/07/2016,00:12:01,
042157/OVERNIGHT/RCPTDISCRP,started,14/07/2016,00:01:42,V0700LBATC
042157/OVERNIGHT/RCPTDISCRP,ended,14/07/2016,00:01:44,
042158/QTCP/QTSMTPCLTP,started,14/07/2016,00:01:53,QSYSWRK
042158/QTCP/QTSMTPCLTP,ended,14/07/2016,01:29:08,
042159/QTCP/QTSMTPCLTP,started,14/07/2016,00:01:53,QSYSWRK
042159/QTCP/QTSMTPCLTP,ended,14/07/2016,00:19:05,
Ctrl+H
Find what: ^([^/]+/[^/]+)/
Replace with: $1,
Replace all
This will replace the second slash of each line by a comma.
You were almost there. You only need to change your replace string with the following:
$1/$2,
How it works
Your regex was: ^(.*?)\/(.*?)\/
In Notepad++'s replace string, the dollar sign is used to refer to groups enclosed by parentheses in the regex.
$1 refers to the first group, (.*?) which is at the beginning of the line, as specified with the ^ character.
$2 refers to the second group, also (.*?), but which follows the first /.
Since you don't want to replace the first slash, you need $1/$2 at the beginning of your replace string. But since what follows the second group is another / (the 2nd one on the line), you need to replace it with the ,. That's why the replace string has to be $1/$2,. Notice that all characters that are not enclosed by ()'s need to be re-written in the replace string. Otherwise, they're just omitted (try replace string $1$2 and you'll see what I mean).
In other editors or programming languages, instead of the $ sign, the \ is sometimes used (sometimes doubled) to refer to parenthetic groups. So you could have for instance \\1/\\2, or \1/\2, as a replace string instead of $1/$2,.
I have something in a text file that looks like '%r'%XXXX, where the XXXX represents some name at the end. Examples include '%r'%addR or '%r'%removeA. I can match these patterns using regex '%r'%\w+. What I would like to replace this with is '{!r}'.format(XXXX). Note that the name has to stay the same in the replace so I'd get '{!r}.format(addR) or '{!r}.format(removeA). Is there a way to replace parts of the matched string in this way while retaining the unknown variable name pulled out with \w+ in the regex search?
I'm specifically looking for a solution using the find and replace features in Notepad++.
You can use
'%r'%(\w+)
and replace with '{!r}.format\(\1\)
The '%r'%(\w+) pattern contains a pair of unescaped parentheses that create a capturing group. Inside the replacement pattern, we use a \1 backreference to restore that value.
NOTE: The ( and ) in the replacement must be escaped because otherwise they are treated as Boost conditional replacement pattern functional characters.
See more on capturing groups and backreferences.
Search on:
'%r'%(XXXX)
Replace with:
Whatever You like \1
\1 will match the first set of grouping parentheses.
I am attempting to edit a csv file, below is a sample line from this file.
|MIGRATE|;|10000|;|2ACC0003|;|30/09/13|;|Positive Adjmt.|;||;|MIGRATE|;|95004U
The beginning of the line |MIGRATE| needs to be modified without changing the second MIGRATE so the line would read
|MIGRATE|;|MIG_IN|;|10000|;|2ACC0003|;|30/09/13|;|Positive Adjmt.|;||;|MIGRATE|;|95004U
There are 7700 or so lines so if I am forced to do this manually I will probably cry a little.
Thanks in advance!
Just replace all the ones you want not changed with another word temporarily, then replace the rest with what you want. I'm not sure what you're asking here, but from what I can guess this might help.
It seems like you could just search for Just search for:
^\|MIGRATE\|
And replace with:
|MIGRATE|;|MIG_IN|
Make sure you've checked 'Regular expression' in the 'Search Mode' options.
Explanation: The ^ is a begin anchor; it will match the beginning of the line, ensuring that it does not match the second |MIGRATE|. The \ characters are required to escape the | characters since they normally have special meaning in regular expressions, and you want to match a literal |.
You can use beginning of line anchors:
Find:
^(\|MIGRATE\|)
Replace with:
$1;|MIG_IN|
regex101 demo
Just make sure that you are using the regular expression mode of the Search&Replace.
If you want to be a bit fancier, you can use a positive lookbehind:
Find:
(?<=^\|MIGRATE\|)
Replace with:
;|MIG_IN|
^ Will match only at the beginning of a line.
( ... ) is called a capture group, and will save the contents of the match in variable you can use (in the first regex, I accessed the variable using $1 in the replace. The first capture gets stored to $1, the second to $2, etc.)
| is a special character meaning 'or' in regex (to match a character or group of characters or another, e.g. a|b matches a or b. As such, you need to escape it with a backslash to make a regex match a literal |.
In my second regex, I used (?<= ... ) which is called a positive lookbehind. It makes sure that the part to be matched has what's inside before it. For instance, (?<=a)b matches a b only if it has an a before it. So that the b in ab matches but not in bb.
The website I linked also explains the details of the regex and you can try out some regex yourself!