I'm trying to add a space between the colon and the character or number if there is no space already:
this is my sentence:hi it's midnight
this is my sentence:2012 was a good year.
this sentence should be ignored: ignore me
And I want it to read:
this is my sentence: hi it's midnight
this is my sentence: 2012 was a good year.
this sentence should be ignored: ignore me
I've tried this with no success.
Regex to insert space in vim
I can get the ":X" or ":0" using this (it doesn't match the space after the colon, which is good):
:[^\s]
But if I replace with:
: $0
It ends up like:
this is my sentence: :hi it's midnight
this is my sentence: :2012 was a good year.
I'm using http://www.gskinner.com/RegExr/ to test the pattern.
I'll be using PHP preg_replace function
Any help is much appreciated.
Thanks in advance.
You can just replace :\s* with :<pretend this is a space>:
preg_replace('/:\s*/', ': ', $text)
:\s* matches : followed by any number of spaces (or no spaces).
Here are two substitution commands to be issued in VIM:
%s/:/:<space>/g
%s/:<space><space>/:<space>/g
Of course, it is an overkill, that is, the number of commands to be issued is two, not one.
Here is two-in-one command:
:%s/:<space>*/:<space>/g
<space> is used to make it more clear as the whitespace inserted by pressing the spacebar is not printable.
Another approach to work around non-printability of space for searching:
:%s/:\s*/:<space>/g
Related
I am using the following regex:
^[0-9.,]*(([.,][-])|([.,][0-9]{2}))?\$
I use this regex to check for valid prices -- so it catches/rejects things like xxx, or llddd or 34.23dsds
and allows things like 100 or 120.00
The problem with it seems to be if it is blank(empty) it passes as valid which it should not -- any ideas how to change this??
Thanks
One of your problems is that you use the dot in your regex which stands for "any character". If you mean a dot you need to escape it like this \.
Also you should have at least one number in it so exchange the asterisk * by a + for "one or more".
Then you can have .,.,.,.,.,.,- if you do not remove the comma and dot from the first part:
^[0-9]+(([\.,][-])|([\.,][0-9]{2}))?$
Taking yoiur regex and just solving the "don't match blanks" problem:
^[0-9.,]+(([.,][-])|([.,][0-9]{2}))?$
the * allows 0 or more, while the + allows 1 or more, thus the * allowed blanks but the + will not, instead there must be at least one digit.
EDIT:
You should clean this regex up a bit to be
^[0-9]+(?:[.,-](?:[0-9]{2})?)?$
This solves the matching of ",,,"
http://www.regextester.com/?fam=95185
EDIT 2: #Fuzzzzel pointed out that this did not match the case "50,-" which we assume you would like to match and that removing capturing groups is presumptive. Here's the latest iteration of my suggested regex:
^[0-9]+([.,-](-|([0-9]{2}))?)?$
I am trying to retrieve matches from a comma separated list that is located inside parenthesis using regular expression. (I also retrieve the version number in the first capture group, though that's not important to this question)
What's worth noting is that the expression should ideally handle all possible cases, where the list could be empty or could have more than 3 entries = 0 or more matches in the second capture group.
The expression I have right now looks like this:
SomeText\/(.*)\s\(((,\s)?([\w\s\.]+))*\)
The string I am testing this on looks like this:
SomeText/1.0.4 (debug, OS X 10.11.2, Macbook Pro Retina)
Result of this is:
1. [6-11] `1.0.4`
2. [32-52] `, Macbook Pro Retina`
3. [32-34] `, `
4. [34-52] `Macbook Pro Retina`
The desired result would look like this:
1. [6-11] `1.0.4`
2. [32-52] `debug`
3. [32-34] `OS X 10.11.2`
4. [34-52] `Macbook Pro Retina`
According to the image above (as far as I can see), the expression should work on the test string. What is the cause of the weird results and how could I improve the expression?
I know there are other ways of solving this problem, but I would like to use a single regular expression if possible. Please don't suggest other options.
When dealing with a varying number of groups, regex ain't the best. Solve it in two steps.
First, break down the statement using a simple regex:
SomeText\/([\d.]*) \(([^)]*)\)
1. [9-14] `1.0.4`
2. [16-55] `debug, OS X 10.11.2, Macbook Pro Retina`
Then just explode the second result by ',' to get your groups.
Probably the \G anchor works best here for binding the match to an entry point. This regex is designed for input that is always similar to the sample that is provided in your question.
(?<=SomeText\/|\G(?!^))[(,]? *\K[^,)(]+
(?<=SomeText\/|\G) the lookbehind is the part where matches should be glued to
\G matches where the previous match ended (?!^) but don't match start
[(,]? *\ matches optional opening parenthesis or comma followed by any amount of space
\K resets beginning of the reported match
[^,)(]+ matches the wanted characters, that are none of ( ) ,
Demo at regex101 (grab matches of $0)
Another idea with use of capture groups.
SomeText\/([^(]*)\(|\G(?!^),? *([^,)]+)
This one without lookbehind is a bit more accurate (it also requires the opening parenthesis), of better performance (needs fewer steps) and probably easier to understand and maintain.
SomeText\/([^(]*)\( the entry anchor and version is captured here to $1
|\G(?!^),? *([^,)]+) or glued to previous match: capture to $2 one or more characters, that are not , ) preceded by optional space or comma.
Another demo at regex101
Actually, stribizhev was close:
(?:SomeText\/([^() ]*)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\))
Just had to make that one class expect at least one match
(?:SomeText\/([0-9.]+)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\)) is a little more clear as long as the version number is always numbers and periods.
I wanted to come up with something more elegant than this (though this does actually work):
SomeText\/(.*)\s\(([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?\)
Obviously, the
([^\,]+)?\,?\s?
is repeated 6 times.
(It can be repeated any number of times and it will work for any number of comma-separated items equal to or below that number of times).
I tried to shorten the long, repetitive list of ([^\,]+)?\,?\s? above to
(?:([^\,]+)\,?\s?)*
but it doesn't work and my knowledge of regex is not currently good enough to say why not.
This should solve your problem. Use the code you already have and add something like this. It will determine where commas are in your string and delete them.
Use trim() to delete white spaces at the start or the end.
$a = strpos($line, ",");
$line = trim(substr($line, 55-$a));
I hope, this helps you!
I just spent hours figuring out how to write a regular expression in PHP that I need to only allow the following format of a string to pass:
(any digit)_(any digit)
which would look like:
219211_2
so far I tried a lot of combinations, I think this one was the closest to the solution:
/(\\d+)(_)(\\d+)/
also if there was a way to limit the range of the last number (the one after the underline) to a certain amount of digits (ex. maximal 12 digits), that would be nice.
I am still learning regular expressions, so any help is greatly appreciated, thanks.
The following:
\d+_\d{1,12}(?!\d)
Will match "anywhere in the string". If you need to have it either "at the start", "at the end" or "this is the whole thing", then you will want to modify it with anchors
^\d+_\d{1,12}(?!d) - must be at the start
\d+_\d{1,12}$ - must be at the end
^\d+_\d{1,12}$ - must be the entire string
demo: http://regex101.com/r/jG0eZ7
Explanation:
\d+ - at least one digit
_ - literal underscore
\d{1,12} - between 1 and 12 digits
(?!\d) - followed by "something that is not a digit" (negative lookahead)
The last thing is important otherwise it will match the first 12 and ignore the 13th. If your number happens to be at the end of the string and you used the form I originally had [^\d] it would fail to match in that specific case.
Thanks to #sln for pointing that out.
You don't need double escaping \\d in PHP.
Use this regex:
"/^(\d+)_(\d{1,12})$/"
\d{1,12} will match 1 to 12 digist
Better to use line start/end anchors to avoid matching unexpected input
Try this:
$regex= '~^/(\d+)_(\d+)$~';
$input= '219211_2';
if (preg_match($regex, $input, $result)) {
print_r($result);
}
Just try with following regex:
^(\d+)_(\d{1,12})$
This is my code to pre_match when an amount looks like this: $ 99.00 and it works
if (preg_match_all('/[$]\s\d+(\.\d+)?/', $tout, $matches))
{ $tot2 = $matches[0];
$tot2 = preg_replace("/\\\$/", '', $tot2);}
I need to do the same thing for a amount that looks like this (with a comma): $ 99,00
Thank you for your help (changing dot for comma do not help, there is an "escape" thing I do not understand...
Idealy I need to preg_match any number that looks like an amount with dot or commas and with or without dollar sign before or after (I know, it's a lot to ask :) since on the result form I want to scan there are phone and street numbers...
UPDATE (For some reason I cannot comment on replies) : To test properly, I need to preg_replace the comma by a dot (since we are dealings with sums, I don't think calculations can be done on numbers with commas in it).
So to clarify my question, I should say : I need to transform, let's say "$ 200,24" to "200.24". (could be amounts bettween 0.10 to 1000.99) :
$tot2 = preg_replace("/\\\$/", '', $tot2);}
(this code just deals with the $ (it works), I need adaptation to deal also with the change of (,) for (.))
No, using , in place of \. works perfectly fine.
It's just that your input does not contain a space between dollar sign and amount $ 99,00 like your .-using source did.
Make the \s optional.
How about:
$str='$ 200,24';
echo str_replace(array('$',',',' '), array('','.',''), $str);
output:
200.24
replace the . character with a character class [,.] which includes both a dot(.) and comma(,)
'/[$]\s\d+([.,]\d+)?/'
edit: comment is correct, regex fixed.
I have a regular expression in PHP that looks for the date in the format of YYYY-MM-DD
What I have is: [\d]{4}-[\d]{2}-[\d]{2}
I'm using preg_match to test the date, the problem is that 2009-11-10 works, but 2009-11-1033434 works as well. It's been awhile since I've done regex, how do I ensure that it stops at the correct spot? I've tried doing /([\d]{4}-[\d]{2}-[\d]{2}){1}/, but it returns the same result.
Any help would be greatly appreciated.
What you need is anchors, specifically ^ and $. The former matches the beginning of the string, the latter matches the end.
The other point I would make is the [] are unnecessary. \d retains its meaning outside of character ranges.
So your regex should look like this: /^\d{4}-\d{2}-\d{2}$/.
^20[0-2][0-9]-((0[1-9])|(1[0-2]))-([0-2][1-9]|3[0-1])$
I added a little extra check to help with the issue of MM and DD getting mixed up by the user. This doesn't catch all date mixups, but does keeps the YYYY part between 2000 and 2029, the MM between 01 and 12 and the DD between 01 and 31
How do you expect your date to be terminated ?
If an end-of-line, then a following $ should do the trick.
If by a non-digit character, then a following negative assertion (?!\d) will similarly work.
you're probably wanting to put anchors on the expression.
i.e.
^[\d]{4}-[\d]{2}-[\d]{2}$
note the caret and dollar sign.
You probably want look ahead assertions (assuming your engine supports them, php/preg/pcre does)
Look ahead assertions (or positive assertions) allow you to say "and it should be followed by X, but X shouldn't be a part of the match). Try the following syntax
\d{4}-\d{2}-\d{2}(?=[^0-9])
The assertion is this part
(?=[^0-9])
It's saying "after my regex, the next character can't be a number"
If that doesn't get you what you want/need, post an example of your input and your PHP code that's not working. Those two items can he hugely useful in debugging these kinds of problems.
[\d]{4}-[\d]{2}-[\d]{2}?
where the question mark means "non-greedy"
You could try putting both a '^' and a '$' symbol at the start and end of your expression:
/^[\d]{4}-[\d]{2}-[\d]{2}$/
which match the start and the end of the string respectively.