there have a long articles, I want only remove thousand separator, not a comma.
$str = "Last month's income is 1,022 yuan, not too bad.";
//=>Last month's income is 1022 yuan, not too bad.
preg_replace('#(\d)\,(\d)#i','???',$str);
How to write the regex patterns? Thanks
If the simplified rule "Match any comma that lies directly between digits" is good enough for you, then
preg_replace('/(?<=\d),(?=\d)/','',$str);
should do.
You could improve it by making sure that exactly three digits follow:
preg_replace('/(?<=\d),(?=\d{3}\b)/','',$str);
If you have a look at the preg_replace documentation you can see that you can write captures back in the replacement string using $n:
preg_replace('#(\d),(\d)#','$1$2',$str);
Note that there is no need to escape the comma, or to use i (as there are not letters in the pattern).
An alternative (and probably more efficient) way is to use lookarounds. These are not included in the match, so they don't have to written back:
preg_replace('#(?<=\d),(?=\d)#','',$str);
The first (\d) is represented by $1, the second (\d) by $2. Therefore the solution is to use something like this:
preg_replace('#(\d)\,(\d)#','$1$2',$str);
Actually it would be better to have 3 numbers behind the comma to avoid causing havoc in lists of numbers:
preg_replace('#(\d)\,(\d{3})#','$1$2',$str);
Related
I want to split a string as per the parameters laid out in the title. I've tried a few different things including using preg_match with not much success so far and I feel like there may be a simpler solution that I haven't clocked on to.
I have a regex that matches the "price" mentioned in the title (see below).
/(?=.)\£(([1-9][0-9]{0,2}(,[0-9]{3})*)|[0-9]+)?(\.[0-9]{1,2})?/
And here are a few example scenarios and what my desired outcome would be:
Example 1:
input: "This string should not split as the only periods that appear are here £19.99 and also at the end."
output: n/a
Example 2:
input: "This string should split right here. As the period is not part of a price or at the end of the string."
output: "This string should split right here"
Example 3:
input: "There is a price in this string £19.99, but it should only split at this point. As I want it to ignore periods in a price"
output: "There is a price in this string £19.99, but it should only split at this point"
I suggest using
preg_split('~\£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})?(*SKIP)(*F)|\.(?!\s*$)~u', $string)
See the regex demo.
The pattern matches your pattern, \£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})? and skips it with (*SKIP)(*F), else, it matches a non-final . with \.(?!\s*$) (even if there is trailing whitespace chars).
If you really only need to split on the first occurrence of the qualifying dot you can use a matching approach:
preg_match('~^((?:\£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})?|[^.])+)\.(.*)~su', $string, $match)
See the regex demo. Here,
^ - matches a string start position
((?:\£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})?|[^.])+) - one or more occurrences of your currency pattern or any one char other than a . char
\. - a . char
(.*) - Group 2: the rest of the string.
To split a text into sentences avoiding the different pitfalls like dots or thousand separators in numbers and some abbreviations (like etc.), the best tool is intlBreakIterator designed to deal with natural language:
$str = 'There is a price in this string £19.99, but it should only split at this point. As I want it to ignore periods in a price';
$si = IntlBreakIterator::createSentenceInstance('en-US');
$si->setText($str);
$si->next();
echo substr($str, 0, $si->current());
IntlBreakIterator::createSentenceInstance returns an iterator that gives the indexes of the different sentences in the string.
It takes in account ?, ! and ... too. In addition to numbers or prices pitfalls, it works also well with this kind of string:
$str = 'John Smith, Jr. was running naked through the garden crying "catch me! catch me!", but no one was chasing him. His psychatre looked at him from the window with a circumspect eye.';
More about rules used by IntlBreakIterator here.
You could simply use this regex:
\.
Since you only have a space after the first sentence (and not a price), this should work just as well, right?
It's a basic preg_replace that detects phone numbers (and just long numbers). My problem is I want to avoid detecting numbers between double "", single '' and forward slashes //
$text = preg_replace("/(\+?[\d-\(\)\s]{8,25}[0-9]?\d)/", "<strong>$1</strong>", $text);
I poked around but nothing is working for me. Your help will be appreciated.
I predict that your pattern is going to let you down more than it is going to satisfy you (or you are very comfortable with "over-matching" within the scope of your project).
While my suggestion really blows out the pattern length, a (*SKIP)(*FAIL) technique will serve you well enough by consuming and discarding the substrings that require disqualification. There may be a way of dictating the pattern logic with lookaround instead, but with an initial pattern with so many potential holes in it and no sample data, there are just too many variables to make a confident suggestion.
Regex101 Demo
Code: (Demo)
$text = <<<TEXT
A number 555555555 then some more text and a quoted number "(123)4567890" and
then 1 2 3 4 6 (54) 3 -2 and forward slashed /+--------0/ versus
+--------0 then something more realistic '234 588 9191' no more text.
This is not closed by the same character on both
ends: "+012345678901/ which of course is a _necessary_ check?
TEXT;
echo preg_replace(
'~([\'"/])\+?[\d()\s-]{8,25}\d{1,2}\1(*SKIP)(*FAIL)|((?!\s)\+?[\d()\s-]{8,25}\d{1,2})~',
"<strong>$2</strong>",
$text);
Output:
A number <strong>555555555</strong> then some more text and a quoted number "(123)4567890" and
then <strong>1 2 3 4 6 (54) 3 -2</strong> and forward slashed /+--------0/ versus
<strong>+--------0</strong> then something more realistic '234 588 9191' no more text.
This is not closed by the same character on both
ends: "<strong>+012345678901</strong>/ which of course is a _necessary_ check?
For the technical breakdown, see the Regex101 link.
Otherwise, this is effectively checking for "phone numbers" (by your initial pattern) and if they are wrapped by ', ", or / then the match is ignored and the regex engine continues looking for matches AFTER that substring. I have added (?!\s) at the start of the second usage of your phone pattern so that leading spaces are omitted from the replacement.
It seems that you're not validating, then you might be trying to write some expression with less boundaries, such as:
^\+?[0-9()\s-]{8,25}[0-9]$
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
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-$a));
I hope, this helps you!
I want to match something like this in PHP:
class-11/xxx/xxx/xx/xxx/things_to_remember/
class-12/xxx/xxx/xx/xxx/things_to_remember/
However I don't want to match something like this:
xxx/class-11/xxx/
class-11/xxx/things_to_remember/xxx
class-11/xxx/
I am writing it like this:
^(class-[12]{2})/.+/things_to_remember/$
I heard regular expression have many features like greedy etc. and they also need to be efficient ? Is the above regualar expression good ?
I wrote a little regex here that captures it like the format you wrote. It might need some slight changes as I didn't know how many digits could be in after class.
/
class-(\d{2}) #Matches class, makes sure that class only is 2 digits - captures class digits
\/([^\/]{3}) #captures first 3 characters that aren't a slash.
\/([^\/]{3}) #Capture 3 characters again.
\/([^\/]{2}) #captures two characters
\/([^\/]{3}) #Captures 3 characters
\/things_to_remember\/ #Matches last piece of string.
/xg
You can test it out here.