I would like to know how can I create a regexp to match the pattern 3,231 or 3,231,201 in php ?
Thanks
/([0-9,]+k?)/
The above regex will match numbers, comma, with an optional 'k' in the end.
A pattern could look like this:
/([0-9]+)(,[0-9]{3})*/
This will allow for something like:
123
123,123
123,123,123
1,123
123456
but not:
123,
123,1
123,12
123,1234
You can modify the behaviour, e.g. allowing for more/less digits after the comma by changing {3} into + or {1,4} (1 to 4) or {3,} (3 or more).
Well that is pretty easy;
/^[0-9,]+$/
this works with (for example) 1231,1231,1312,12312
Related
I have these two regular expression
^(((98)|(\+98)|(0098)|0)(9){1}[0-9]{9})+$
^(9){1}[0-9]{9}+$
How can I combine these phrases together?
valid phone :
just start with : 0098 , +98 , 98 , 09 and 9
sample :
00989151855454
+989151855454
989151855454
09151855454
9151855454
You haven't provided what passes and what doesn't, but I think this will work if I understand correctly...
/^\+?0{0,2}98?/
Live demo
^ Matches the start of the string
\+? Matches 0 or 1 plus symbols (the backslash is to escape)
0{0,2} Matches between 0 and 2 (0, 1, and 2) of the 0 character
9 Matches a literal 9
8? Matches 0 or 1 of the literal 8 characters
Looking at your second regex, it looks like you want to make the first part ((98)|(\+98)|(0098)|0) in your first regex optional. Just make it optional by putting ? after it and it will allow the numbers allowed by second regex. Change this,
^(((98)|(\+98)|(0098)|0)(9){1}[0-9]{9})+$
to,
^(?:98|\+98|0098|0)?9[0-9]{9}$
^ this makes the non-grouping pattern optional which contains various alternations you want to allow.
I've made few more corrections in the regex. Use of {1} is redundant as that's the default behavior of a character, with or without it. and you don't need to unnecessarily group regex unless you need the groups. And I've removed the outer most parenthesis and + after it as that is not needed.
Demo
This regex
^(?:98|\+98|0098|0)?9[0-9]{9}$
matches
00989151855454
+989151855454
989151855454
09151855454
9151855454
Demo: https://regex101.com/r/VFc4pK/1/
However note that you are requiring to have a 9 as first digit after the country code or 0.
Hi I tired to use RegEx in PHP. The following elements I like to get with it:
<a="300">
<a="300"b="300">
<b="300">
The Problem is that I get only
<a="300">
<b="300">
with the following RegEx:
<(a|b)="[0-9]*">
What do I have to change, that I get all three elements? Is there a ANDOR operator?
Assuming your problem is rather a simple string processing than serious parsing, I would modify your regex like this:
<(a|b)="[0-9]+".*>
I added .* to allow characters inbetween " and >.
or a slightly my-flavored version:
<[ab]="\d+"[^>]*?>
piping single characters with | are less favored over [...]
\d is for series of digits
[^>]*? for characters other than >
You need an additional grouping, to specify, that you would accept multiple of that kind:
echo '<a="300">
<a="300"b="300">
<b="300">' | egrep '<((a|b)="[0-9]*")+>'
<a="300">
<a="300"b="300">
<b="300">
Regex is not boolean logic. The | symbol in regex is not an OR operator; it is referred to as alternation, which works similarly but is not quite the same thing. If you're just trying to match one of multiple characters, you should use square brackets [] to create a character set. In this case, [ab] matches a or b, just as [0-9] matches 0 or 1 or 2 etc.
Here's the pattern that I would suggest
<[ab]="\d+"(?:[ab]="\d+")?>
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.
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})$
I have the following string that I need to match only the last seven digets between [] brackets. The string looks like this
[15211Z: 2012-09-12] ([5202900])
I only need to match 5202900 in the string contained between ([]), a similar number could appear anywhere in the string so something like this won't work (\d{7})
I also tried the following regex
([[0-9]{1,7}])
but this includes the [] in the string?
If you just want the 7 digits, not the brackets, but want to make sure that the digits are surrounded with brackets:
(?<=\[)\d{7}(?=\])
FYI: This is called a positive lookahead and positive lookbehind.
Good source on the topic: http://www.regular-expressions.info/lookaround.html
Try matching \(\[(\d{7})\]\), so you match this whole regular expression, then you take group 1, the one between unescaped parentheses. You can replace {7} with a '*' for zero or more, + for 1 or more or a precise range like you already showed in your question.
You can try to use
\[(\d{1,7})\]
If first pattern looks like yours (not only digits), then this should work for you to extract group of digits surrounded by brackets like ([123]):
\(\[(\d+)\]\)
From your details, lookbehind and lookaround seems to be good one. You can also use this one:
(\d{7})\]\)$
Since the pattern of seven digit is expected at the end of the line, engine need to work less in order to find the match.
Hope it helps!
Here is a benchmark (in Perl, but I think is close the same in php) that compares lookaround approach and capture group:
use Benchmark qw(:all);
my $str = q/[15211Z: 2012-09-12] ([5202900])/;
my $count = -3;
cmpthese($count, {
'lookaround' => sub {
$str =~ /(?<=\[)\d{7}(?=\])/;
},
'capture group' => sub {
$str =~ /\[(\d{7})\]/;
},
});
result:
Rate lookaround capture group
lookaround 274914/s -- -70%
capture group 931043/s 239% --
As we can see, capture is more than 3 times faster than lookaround.