I am using PHP.
I have a strings like:
example.123.somethingelse
example.1234.somethingelse
example.2015.123.somethingelse
example.2015.1234.somethingelse
and I came up with this regex
/example\.(2015\.|)([0-9]{3,4})\./
What I want to get is "123" or "1234" and it works for these strings. But when the string is
example.2015.A01.somethingelse
the result is "2015".
The way that I see it, after "2015." I have "A" and this should not be matched by the regex, but it is ( and I suppose there is a solid reason for it that I dont understand atm).
How can I fix it ( make the regex match nothing since the last string does not follow the same structure as the others) ?
Your regex is this:
/example\.(2015\.|)([0-9]{3,4})\./
That says
First match "example" followed by a period
Then match either "2015" followed by a period OR nothing at all.
Then match 3 or 4 digits in a row followed by a period
When you have the string example.2015.A01.somethingelse it matches the "example.2015." but then, as you said, the "A" messes it up so it backtracks and matches just "example." (remember the "OR" allowed for nothing to be matched). So it matches "example." followed by NOTHING followed by 3 or 4 numeric digits -- since "2015" is 4 numeric digits it comfortably matches "example.2015".
It's hard to tell from your description, but I think you've just got a mis-placed vertical bar:
/example\.(2015\.)|([0-9]{3,4})\./
That should match EITHER "example.2015." OR numbers like 123 -- but "2015" is still 4 numeric digits in a row, so it will still match. I don't have a clear enough idea of the pattern to figure out how that could be avoided.
Maybe use \d+ and take the first result in the array.
In your regex, you use the following:
(2015\.|)
This allows the regex to match either 2015. or the empty string (zero characters).
When the regex example\.(2015\.|)([0-9]{3,4})\. is applied to the following example:
example.2015.A01.somethingelse
it will to match the literal characters example, and then the empty string with (2015\.|) and then uses ([0-9]{3,4})\. to match the string 2015, which is 4 numerical characters. Thus your expression matches the following:
example.2015.
Looks like you need a possessive quantifier:
/example\.(2015\.)?+([0-9]{3,4})\./
The 2015. is still optional, but once the regex has matched it, it won't give it up, even if that causes the match to fail. I'm assuming the substring you're trying to capture with ([0-9]{3,4}) can never have the value 2015. That is, you won't need to match something like this:
example.2015.somethingelse
If that's not the case, it's going to be much more complicated.
here is one more pattern
example\.(?:2015\.)?\K(\d+)
Demo
or to your specific amount of digits
example\.(?:2015\.)?\K(\d{3,4})
Related
I need a regular expression to detect the phrase Figure 1.5: in a given string. Also, I intend on using this expression in a PHP preg_replace() function.
Here are some more examples:
...are given. Figure 2.1: shows that...
...are given. Figure 3: shows that...
...are given. Figure 1.16: shows that...
...are given. Figure 0.4 shows that...
...are given. figure 5.1: shows that...
With my limited Regex knowledge, I was able to create this:
/\wFigure \d*\.?\d*/g
But that doesn't even begin to handle all of the permutations that could occur.
I would appreciate any suggestions that you might have.
There are several points here:
You are using \w at the start, perhaps, as a word boundary. In fact, \w matches a letter, digit or _ and actually requires this char to be at the exact location. However, there is no word char before Figure, so you need to either remove \w or replace with \b.
preg_replace replaces all non-overlapping occurrences by default, you do not need the g modifier
\d*\.?\d* is fine here, but since you want to match any digits followed with zero or more occurrences of . and digits you can use a more specific pattern like \d+(?:\.\d+)*.
You can use
preg_replace('/Figure \d+(?:\.\d+)*/', '', $string)
See the regex demo.
Details:
Figure - a string
- a space (replace with \s+ to match any one or more whitespaces, and consider adding u flag after last / if you need to find all Unicode whitespaces)
\d+ - one or more digits
(?:\.\d+)* - zero or more occurrences of . and one or more digits.
I want to extract an 8 digit number from a string using regex.
Test strings are:
hi 82799162
236232 (82342450)
test data 8979
Required respective output should be
82799162
82342450
null
I have tried following code:
preg_match('/[0-9]{8}$/', $string, $match);
preg_match('/\d{8}$/', $string, $match);
But neither retrieves the number from 236232 (82342450).
If a regex is to capture exactly 8 digits, is must contain:
\d{8} as a central part,
a "before" condition, ensuring that no digit occurs before your match,
an "after" condition, ensuring that no digit occurs after your match.
One of possible solutions is to use negative lookbehind / lookahead:
(?<!\d)\d{8}(?!\d)
Another option is word boundary assertions (at both ends):
\b\d{8}\b
I think, regex like [0-9]{8} is not enough, as it captures also
first 8 digits from a longer sequence of digits.
Are you happy with that?
The problem is with your $ sign, and it is used to indicate the end of your expression. So basically, with that expression, you are looking for a string which ends with a 8 digit number. But in your second test string; '236232 (82342450)', ends with a bracket, and therefore it doesn't match the criteria (does not end with a number).
So remove the trailing $ and it will work.
preg_match('/[0-9]{8}/',$string,$match);
Hope it helps!!
I have user-entered text with potentially mistyped "tokens" I'm trying to find using PHP.
A valid "token" is any number of word characters wrapped in percent signs - so %blah% %blah_moreblah%. Basically I'm looking for tokens where the user may have forgotten to put a leading or trailing '%'. I'm also looking for tokens in the valid format - as at this point in my code, all replaceable tokens have already been replaced.
So, the 3 situations I'm looking for are (to borrow regex syntax): %\w+, %\w+%, or \w+%.
In English, what I'm looking for is, "a string that starts with a % and/or ends with a % and contains only word characters'
The regex I have this far is: (%*\w+%*), but you'll notice it matches every single word. I'm stuck on making a match require at least a leading or a trailing %.
Edit: Initially I tried to have all 3 situations found with their own regex. However, I was finding that the regex for finding tokens in the first situation would also find tokens in the second situation, just without the trailing %. For example, /(%\w+)/, when checked against %before %both%, would match %before and %both.
To match tokens enclosed with %, or having % on either side, use
(?=\w*%)%*\w+%*
See another regex demo.
This is your pattern that I added a positive lookahead to. The (?=\w*%) restricts to only such matches where a % appears after a zero or more occurrences of word characters.
Note also that %* will match zero or more percent signs, it may match %%%word%%. If it is not what you need, and if you need to match 1 or 0 %s, just replace the * with ? quantifier.
Try this:
$input_lines = "Hello this is a %string% with %some_words in it just for demo% purposes.";
preg_match_all("/\s[\w_\-]+%\.?|%[\w_\-]+(%|\s|\.)/", $input_lines, $output_array);
That will output this:
array(
0 => %string%
1 => %some_words
2 => demo%
)
Note that this will catch the valid cases, as well as the typos you are looking for.
I've got a scenario as follows. Our systems needs to pull filters from a string passed in as a query parameter, but also throw a 404 error if the string isn't correctly formatted. So let's take the following three strings as an exmple:
pf0pt1000r
pfasdfadf
pf2000pt2100
By the application requirements, only #3 is supposed to match as a "valid" string. So my current regex to match that is /([a-z]+)(\d+)/. But this also matches #1, if not entirely, but it still matches.
My problem thus is twofold - I need 2 patterns, 1 that will match only the 3rd string in this list, and another that will match the "not-acceptable" strings 1 and 2. I believe there must be some way to "negate" a pattern (so then I'd technically only need one pattern, I'm assuming), but I'm not sure how exactly to do that.
Thanks for any help!
EDIT
For clarity's sake, let me explain. The "filter parameters" present here take the following structure - 1 or 2 letters, followed by a number of, well, numbers. That structure can repeat itself however many times. So for example, valid filter strings could be
pf100pt2000
pf100pt2000r2wp0to1
etc.
Invalid strings could be
pf10000pt2000r
pf3000pt2123wpno
... anything not following the structure above.
After clarifying the question:
^([a-zA-Z]{1,2}\d+)*$
Explanation:
[a-zA-Z] - a lower or upper case letter
{1,2} - one or two of those
\d+ - one or more digits
()* - the whole thing repeated any number of times
^$ - match the entire string from start(^) to end($)
You can use this regex for valid input:
^([a-zA-Z]+\d+)+$
RegEx Demo 1
To find invalid inputs use:
^(?!([a-zA-Z]+\d+)+$).+$
RegEx Demo 2
/^(?:(?:[a-z]+)(?:\d+))*$/
You were hella close, man. Just need to repeat that pattern over and over again till the end.
Change the * to a + to reject the empty string.
Oh, you had more specific requirements, try this:
/^(?:[a-z]{1,2}\d+)*$/
Broken down:
^ - Matches the start of the string an anchor
(?: - start a non-capturing group
[a-z] - A to Z. This you already had.
{1,2} - Repeat 1 or 2 times
\d+ - a digit or more You had this, too.
)* - Repeat that group ad nauseum
$ - Match the end of the string
If you only want digits at the end of the string, then
/\d$/
would do. \d = digit, $ = end of string.
I have to create regex to match ugly abbreviations and numbers. These can be one of following "formats":
1) [any alphabet char length of 1 char][0-9]
2) [double][whitespace][2-3 length of any alphabet char]
I tried to match double:
preg_match("/^-?(?:\d+|\d*\.\d+)$/", $source, $matches);
But I coldn't get it to select following example: 1.1 AA My test title. What is wrong with my regex and how can I add those others to my regex too?
In your regex you say "start of string, followed by maybe a - followed by at least one digit or followed by 0 or more digits, followed by a dot and followed by at least one digit and followed by the end of string.
So you regex could match for example.. 4.5, -.1 etc. This is exactly what you tell it to do.
You test input string does not match since there are other characters present after the number 1.1 and even if it somehow magically matched your "double" matching regex is wrong.
For a double without scientific notation you usually use this regex :
[-+]?\b[0-9]+(\.[0-9]+)?\b
Now that we have this out of our way we need a whitespace \s and
[2-3 length of alphabet]
Now I have no idea what [2-3 length of alphabet] means but by combining the above you get a regex like this :
[-+]?\b[0-9]+(\.[0-9]+)?\b\s[2-3 length of alphabet]
You can also place anchors ^$ if you want the string to match entirely :
^[-+]?\b[0-9]+(\.[0-9]+)?\b\s[2-3 length of alphabet]$
Feel free to ask if you are stuck! :)
I see multiple issues with your regex:
You try to match the whole string (as a number) by the anchors: ^ at the beginning and $ at the end. If you don't want that, remove those.
The number group is non-catching. It will be checked for matches, but those won't be added to $matches. That's because of the ?: internal options you set in (?:...). Remove ?: to make that group catching.
You place the shorter digit-pattern before the longer one. If you swap the order, the regex engine will look for it first and on success prefer it over the shorter one.
Maybe this already solves your issue:
preg_match("/-?(\d*\.\d+|\d+)/", $source, $matches);
Demo