I need to check input with preg_match, which must be of this format: xxx.xxx.xxx
The number of block can vary... These are all examples of valid inputs:
001
00a.00a
0fg.001
aaa.aaa.001
001.001.002.001.001.001
Well I could probably write a regexp something like:
^([\da-z]{3}\.?)+$
But here comes the problem with the quantifier of the period. I mean if I use '?' to match 0 or 1 times, it would also match even if skip the dots somewhere, eg:
000.001.0010az001
then, if I used {1} to match one time, it would match nothing, because the last block does not have a dot.
So I can't think what to think of... Please advice
You can use:
^[\da-z]{3}(?:\.[\da-z]{3})*$
/^(?:[\da-z]{3}\.)*[\da-z]{3}$/
Related
$my_string = '88888805';
echo preg_replace("/(^.|.$)(*SKIP)(*F)|(.)/","*",$,my_string);
This shows the first and last number like thus 8******5
But how can i show this number like this 888888**. (The last 2 number is hidden)
Thank you!
From this: 8******5
To: 888888**
I'm not sure if you have worked on this Regex pattern to do something unique. However, I will provide you with a general one that should fit your question without using your current pattern.
$my_string = '88888805';
echo preg_replace("/([0-9]+)[0-9]{2}$/","$1**",$,my_string);
Explanation:
The ([0-9]+) will match all digits, this could be replaced with \d+, it's between brackets to be captured as we are going to use it in the results.
[0-9]{2} is going to match the last 2 digits, again, it can be replaced with \d{2}, it's outside the brackets because we don't want to include them in the result. the $ after that is to indicate the end of the test, it's optional anyways.
Results:
Input: 88888805
Output: 888888**
echo preg_replace("/(.{2}$)(*SKIP)(*F)|(.)/","*",$my_string);
If it for a uni assignment, you'd probably want to do this. Basically says, don't match if its the last two characters, otherwise match.
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 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})$
My current one works almost perfectly but it misses out any other digits after a character like - or /.
The Original String is something like:
#!012 , #!02/09#!011 #!04-072
My current one works on stuff like:
$viewedResult = preg_replace('/#!([0-9A-Z]{1,4})/', '${1}', $viewedResult);
This would give me something like:
04-072<br />
but I want the "-072" in the Number2 bit like:
04-072
It could also be able to include /072s:
04/072
Any ideas? Remember that there is a #! in front of the number!
You could use something like:
preg_replace('~#!([0-9A-Z][0-9A-Z/-]{0,5})~', '${1}', $viewedResult);
I would split up the regular expression in two, as I think you want to match something like \d{1..4}[-/]?\d{1..4} where the first and second set of digits total 4. It's probably better just to look for (\d+[-/]/d+) and do a check afterwards if the total digits size is correct, e.g. finding all matches of (\d+) in the result, and programmatically check their length.
In general, I would pay close attention to what exactly is allowed input, and what (error) you will get if the input does not match. Splitting things up will make it much easier to show a correct error message, such as "number of digits in mooring spot incorrect" (which is I presume what this is about).
If you want the value of Number2, then use:
Number2\s*=\s*([^;"]*)[;"]
And then use the first capture group.
Stupid question: why not using this regular expression /#!([0-9A-Z/-]{1,4})/ ?
Side note: the {1,4} blocks you will retrieving the value 04-072 (it's 6 chars long).
Hope that helps :)
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.