First of all, regular expressions are quite possibly the most confusing thing I have every dealt with - with that being said I cannot believe how efficient they can make ones life.
So I am trying to understand the wildcard regex with no luck
Need to turn
f_firstname
f_lastname
f_dob
f_origincountry
f_landing
Into
':f_firstname'=>$f_firstname,
':f_lastname'=>$f_lastname,
':f_dob'=>$f_dob,
':f_origincountry'=>$f_origincountry,
':f_landing'=>$f_landing,
In the answer can you please briefly describe the regex you are using, I have been reading the tutorials but they boggle my mind. Thanks.
Edit: As Chris points out, you can improve the regex by cleaning up any white space there may be in the target string. I also replace the dot with \w as he did because it's better practice than using the .
Search: ^f_(\w+)\s*$
^ # start at the beginning of the line
f_ # look for f_
(\w+) # capture in a group all characters
\s* # optionally skip over (don't capture) optional whitespace
$ # end of the line
Replace: ':f_\1'=>$f_\1,
':f_ # beginning of replacement string
\1 # the group of characters captured above
'=>$f_ # some more characters for the replace
\1, # the capture group (again)
Find: (^.*)
Replace with: ':$1'=>$$1,
Find What:
(f_\w+)
Here we're matching f_ followed by a word character \w+ (the plus mean one or more times). Wrapping the whole thing in brackets means we can reference this group in the replace pattern
Replace With:
':\1'=>$\1,
This is simply your result phrase but instead of hardcoding the f words I've put \1 to reference the group in the search
Related
I am trying to use regex to match anything but "id":digits part
I have come up with this "(\b(id":)(\d+)\b)" to find the id:byDigits pattern, but I need to negate that but haven't been able to get around it.
[{"age":1,"id":123,"value":"14"},
{"age":1,"id":4214,"value":"4324"},
{"age":3,"id":4244,"value":"545"}]
Any help is appreciated.
Simplest option is to capture the rest of the string into groups and use it in the substituion as below
Demo: https://regex101.com/r/cRVA5C/2/
Pattern: ^([\s\S]*?)\s*"id":\d+,?\s*([\s\S]*?)$
Breakdown:
([\s\S]*?): match any number of any characters before and after "id":. Capture it into groups \1 and \2
\s*"id":\d+,?\s*: match "id"=\d+, optionally preceded by spaces and optionally followed by spaces and ,.
In substituition, use \1\2, to get the desired output.
Note: Regex may not be the ideal tool for parsing JSON.
In my LaTeX work I need to do Regex search with \|(.*?)\| to capture |whatever| and replace it with \somecommand{$1}. But I do not want to capture || (That is, there is nothing between them.) How should I refine my regex search?
(By the way, what should my title be, so that it is useful for others?)
Change your regex to,
\|[^|]+\|
OR
\|.+\|
If you want to also capture pipes in between searched content
You have to change the asterix (which matches 0+ times) to a plus sign make the quantifier match at least 1 character.
\|(.+?)\|
^
I'm trying to find all lines without ending period (dot) but without finding blank (empty) lines. And after that I want to add ending period to that sentence.
Example:
The good is whatever stops such things from happening.
Meaning as the Higher Good
It was from this that I drew my fundamental moral conclusions.
I have tried few regex but they also find empty lines as well.
Is there a regex for Notepad++ that can achieve that?
Enable Regular Expression match, then search for:
\S(?<!\.)\K\s*$
and replace with:
.$0
Breakdown:
\S Match a non-whitespace character
(?<!\.) It shouldn't be a period
\K Reset match
\s* Match optional whitespace characters
$ End of line
You could use something like this to find the lines that you are interested in adding capture group to it and appending you needed chars.
(?<!\.)\r\n
This works by using negative look behind (?<!\.) to check that there is no . before \r
There is a group or regex operators that can be used to accomplish this type of tasks.
Look ahead positive (?=)
Look ahead negative (?!)
Look behind positive (?<=)
Look behind negative (?
Try this short and effective solution too.
Search: \w$
Replace: $0.
In Notepad++, I use the expression (?<=").*(?=") to find all strings in between quotes. It would the seem rather trivial to be able to only keep those results. However, I cannot find an easy solution for this.
I think the problem is that Notepad++ is not able to make multiple selections. But there must be some kind of workaround, right? Perhaps I must invert the regex and then find/replace those results to end up with the strings I want.
For example:
blablabla "Important" blabla
blabla "Again important" blablabla
I want to keep:
Important
Again important
There is no great solution for this and depending on your use case I would recommend writing a quick script that actually uses your first expression and creates a new file with all of the matches (or something like this). However, if you just want something quick and dirty, this expression should get you started:
[^"]*(?:"([^"]*)")?
\1\n
Explanation:
[^"]* # 0+ non-" characters
(?: # Start non-capturing group
" # " literally
( # Start capturing group
[^"]* # 0+ non-" characters
) # End capturing group
" # " literally
)? # End non-capturing group AND make it optional
The reason the optional non-capturing group is used is because the end of your file may very well not have a string in quotes, so this isn't a necessary match (we're more interested in the first [^"]* that we want to remove).
Try something like this:
[^"\r\n]+"([^"]+)"[^"\r\n]+
And replace with $1. The above regex assumes there will be only 2 double quotes in each line.
[^"]+ matches non-quote characters.
[^"\r\n]+ matches non-quote, non newline characters.
regex101 demo
Hard to be certain from your post, but I think you may want : SEE BELOW
<(?<=")(.*)(?=")
The part you keep will be captured as \2.
(?<=")(.*)(?=")
\1 \2 \3
Your original regex string uses parentheses to group characters for evaluation. Parentheses ALSO group characters for capturing. That is what I added.
Update:
The regex pattern you provided doesn't seem to work correctly. Won't this work?
\"(.*)\"
\1 now captures the content.
Is it possible for a regex to match based on other parts of the same regex?
For example, how would I match lines that begins and end with the same sequence of 3 characters, regardless of what the characters are?
Matches:
abcabc
xyz abc xyz
Doesn't Match:
abc123
Undefined: (Can match or not, whichever is easiest)
ababa
a
Ideally, I'd like something in the perl regex flavor. If that's not possible, I'd be interested to know if there are any flavors that can do it.
Use capture groups and backreferences.
/^(.{3}).*\1$/
The \1 refers back to whatever is matched by the contents of the first capture group (the contents of the ()). Regexes in most languages allow something like this.
You need backreferences. The idea is to use a capturing group for the first bit, and then refer back to it when you're trying to match the last bit. Here's an example of matching a pair of HTML start and end tags (from the link given earlier):
<([A-Z][A-Z0-9]*)\b[^>]*>.*?</\1>
This regex contains only one pair of parentheses, which capture the string matched by [A-Z][A-Z0-9]* into the first backreference. This backreference is reused with \1 (backslash one). The / before it is simply the forward slash in the closing HTML tag that we are trying to match.
Applying this to your case:
/^(.{3}).*\1$/
(Yes, that's the regex that Brian Carper posted. There just aren't that many ways to do this.)
A detailed explanation for posterity's sake (please don't be insulted if it's beneath you):
^ matches the start of the line.
(.{3}) grabs three characters of any type and saves them in a group for later reference.
.* matches anything for as long as possible. (You don't care what's in the middle of the line.)
\1 matches the group that was captured in step 2.
$ matches the end of the line.
For the same characters at the beginning and end:
/^(.{3}).*\1$/
This is a backreference.
This works:
my $test = 'abcabc';
print $test =~ m/^([a-z]{3}).*(\1)$/;
For matching the beginning and the end you should add ^ and $ anchors.