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.
Related
In a regular expression (notepad++), I want to search for:( )|(:)|(_)|(\.), and to insert \ before to, as above, a blank space, colon, under line and ".".
Search example: abcd:1234 jiod.8ufd_adfd
Result: abcd\:1234\ jiod\.8ufd\_adfd
Briefly, how can I refer to what was found in the replace expression?
Note that it is not \1, \2, \3 or \4 in the example, as I need to include what was found, there is no way to know which was found, is there?
You can use a single character class (instead of using the alternation with capturing groups) to match one of the listed
In the replacement use $& to refer to the matched text and prepend a backslash.
Match
[:\h._]
Replace with
\\$&
The character class matches either a colon, horizontal whitespace char, dot or underscore.
Regex demo
There's no such thing as insert, because if you think about it, inserting is just replacing the original with a new string that contains the old text as well.
Try this instead: search for ([ :_.]) (your original regex is pointlessly complicated) and replace with \\$1 (ie, slash followed by the original text).
I am using the Kate editor. Here is a minimal example that shows my problem:
I have a file with a bunch of occurrences of:
\command{stuff}
where stuff is some arbitrary string of letters. I want to replace this with
\disobey{stuff}
where stuff is unchanged. The regular expression:
\\command\{[a-zA-Z]*\}
matches such expressions. So I pull the replace dialog with CTRL-r, and enter
Find: \\command\{[a-zA-Z]*\}
Replace: \\disobey\{\1\}
So in the document, an actual instance is say,
\command{exchange}
and when I hit the replace button is changed to
\disobey{1}
In the Kate documentation: Appendix B: Regular Expressions, \1 should match the first pattern used. Is this indeed the correct syntax? I have also tried $1, #1, and various other things.
Here is a quote directly from the documentation:
The string \1 references the first sub pattern enclosed in parentheses
So you need to put [a-zA-Z]* in a capturing group, like ([a-zA-Z]*).
Find: \\command\{([a-zA-Z]*)\}
Replace: \\disobey\{\1\}
Wrap the value with ( ) to capture it as a group, so you can use it in your replace
So change your find regex like this:
\\command\{([a-zA-Z]*)\}
and you should do fine.
I want to use a regular expression in VBA to do a search, capture part of it, and then use that part in the replacement. For example, I want to run the search and replace on these lines:
(a4a)
(aHa)
And get a result of:
(b4b)
(bHb)
How do I capture the 2nd character and use it again in the replacement?
In VBA, capture parts of the search with parenthesis () and use them in replacement with $ and the number of the capture occurrence. Note, normal parenthesis need to be escaped, which is the opposite of vim.
So in this case:
searchPattern = "\(a(.)a\)"
replacement = "(b$1b)"
I'd like to replace this
y[100] with this Ith(y,100) in Sublime3.
I've got the regular expression \by[\d+] in find what and Ith(y,$1) in replace with, but it doesn't work. It finds what to replace correctly but just replaces it with Ith(y, )
You need to pick the \d+ inside () to capture in group($1)
\by[(\d+)]
Also, you may need to to escape the [ and ] characters here.
\by\[(\d+)\]
You have to capture the data you want to backreference: to do that you have to use unescaped parenthesis (...), ie capturing groups. Also, [...] are character classes, a special character in regex that you need to escape.
Try replacing
\by\[(\d+)\]
with
Ith(y, \1)
You need to escape the special characters and capture the number in a capturing group group. Try this regex:
\by\[(\d+)]\b
And replace with
Ith(y, $1)
Online explanation and demonstration: http://regex101.com/r/nX3yJ9
grep "http:\/\/.*\.jpg" index.html -o
Gives me text starting with http:// and ending with .jpg
So does: grep "http:\/\/.*\.\(jpg\)" index.html -o
What is the difference? And is there any condition where this might fail?
I got it to match either jpg,png or gif using this regex:
http:\/\/.*\.\(jpg\|png\|gif\)
Something to do with backreference or regex grouping that I read. Cannot understand this part \(\)
Grouping is used for two purposes in regular expressions.
One uses is to delimit parts of the regexp when using alternatives. That's the case in your third regexp, it allows you to say that the extension can be any of jpg, png, or gif.
The other use is for backreferences. This allows you to refer to the text that matched an earlier part of the regexp later in the regexp. For instance, the following regexp matches any letter that appears twice in a row:
\([a-z]\)\1
The backreference \1 means "match whatever matched the first group in the regexp".
( and ) are metacharacters. i.e. they don't match themselves, but mean something to grep.
From here:
Grouping is performed with backslashes followed by parentheses ‘(’,
‘)’.
so in the above the \( and \) define within them a group of possibilities to match separated by the | character. i.e. your filename extensions.