The following is in PHP but the regex will also be used in javascript.
Trying to extract repeating patterns from a string
string can be any of the following:
"something arbitrary"
"D123"
"D111|something"
"D197|what.org|when.net"
"D297|who.197d234.whatever|when.net|some other arbitrary string"
I'm currently using the following regex: /^D([0-9]{3})(?:\|([^\|]+))*/
This correctly does not match the first string, matches the second and third correctly. The problem is the third and fourth only match the Dxxx and the last string. I need each of the strings between the '|' to be matched.
I'm hoping to use a regex as it makes it a single step. I realize I could just detect the leading Dxxx then use explode or split as appropriate to break the strings out. I've just gotten stuck on wanting a single regular expression match step.
This same regex may be used in Python as well so just want a generic regex solution.
There is no way to have a dynamic number of capture groups in a regular expression, but if you know some upper limit to how many parts you would have in one string, you can just repeat the pattern that many times:
/^D([0-9]{3})(?:$|\|)(.*?)(?:$|\|)(.*?)(?:$|\|)(.*?)(?:$|\|)(.*?)(?:$|\|)/
So after the initial ^D([0-9]{3})(?:$|\|) you just repeat (.*?)(?:$|\|) as many times as you need it.
When the string has fewer elements, those remaining capture groups will match the empty string.
See regex tester.
Is something like preg_match_all() (the PHP variant of a global match) also acceptable for you?
Then you could use:
^(?|D([0-9]{3})|^.+$|(?!^)\|([^|\n]*)(?=\||$))
This will match everything in a string in different matches, e.g. take your string:
D197|what.org|when.net
It will you then give three matches:
D197
what.org
when.net
Running live: https://regex101.com/r/jL2oX6/4 (Everything in green are your group matches. Ignore what's in blue.)
Related
EDIT: the first regex is used only to validate the input. The second one is then used to tell if we should disregard the first part (using split by \s) or not - hence the change to startsWith.
I have inherited code that has to check codes that match a specific regex:
[A-Za-z0-9 -]+(?:[A-Za-z0-9*])$
I changed this to remove the regex group.
Now I would like to test that is the same as the regex above.
[A-Za-z0-9 -]+[A-Za-z0-9*]$
how would I test this without running all the permutations in the world?
I am trying to regex the following string:
https://www.amazon.com/Tapps-Top-Apps-and-Games/dp/B00VU2BZRO/ref=sr_1_3?ie=UTF8&qid=1527813329&sr=8-3&keywords=poop
I want only B00VU2BZRO.
This substring is always going to be a 10 characters, alphanumeric, preceded by dp/.
So far I have the following regex:
[d][p][\/][0-9B][0-9A-Z]{9}
This matches dp/B00VU2BZRO
I want to match only B00VU2BZRO with no dp/
How do I regex this?
Here is one regex option which would produce an exact match of what you want:
(?<=dp\/)(.*)(?=\/)
Demo
Note that this solution makes no assumptions about the length of the path fragment occurring after dp/. If you want to match a certain number of characters, replace (.*) with (.{10}), for example.
Depending on your language/method of application, you have a couple of options.
Positive look behind. This will make your regex more complicated, but will make it match what you want exactly:
(<=dp/)[0-9A-Z]{10}
The construct (<=...) is called a positive look behind. It will not consume any of the string, but will only allow the match to happen if the pattern between the parens is matched.
Capture group. This will make the regex itself slightly simpler, but will add a step to the extraction process:
dp/([0-9A-Z]{10})
Anything between plain parens is a capture group. The entire pattern will be matched, including dp/, but most languages will give you a way of extracting the portion you are interested in.
Depending on your language, you may need to escape the forward slash (/).
As an aside, you never need to create a character class for single characters: [d][p][\/] can equally well be written as just dp\/.
As the title says, I'm trying to build up a regular expression that can recognize strings with this format:
word!!cat!!DOG!! ... Phone!!home!!
where !! is used as a delimiter. Each word must have a length between 1 and 5 characters. Empty words are not allowed, i.e. no strings like !!,!!!! etc.
A word can only contain alphabetical characters between a and z (case insensitive). After each word I expect to find the special delimiter !!.
I came up with the solution below but since I need to add other controls (e.g. words can contain spaces) I would like to know if I'm on the right way.
(([a-zA-Z]{1,5})([!]{2}))+
Also note that empty strings are not allowed, hence the use of +
Help and advices are very welcome since I just started learning how to build regular expressions. I run some tests using http://regexr.com/ and it seems to be okay but I want to be sure. Thank you!
Examples that shouldn't match:
a!!b!!aaaaaa!!
a123!!b!!c!!
aAaa!!bbb
aAaa!!bbb!
Splitting the string and using the values between the !!
It depends on what you want to do with the regular expression. If you want to match the values between the !!, here are two ways:
Matching with groups
([^!]+)!!
[^!]+ requires at least 1 character other than !
!! instead of [!]{2} because it is the same but much more readable
Matching with lookahead
If you only want to match the actual word (and not the two !), you can do this by using a positive lookahead:
[^!]+(?=!!)
(?=) is a positive lookahead. It requires everything inside, i.e. here !!, to be directly after the previous match. It however won't be in the resulting match.
Here is a live example.
Validating the string
If you however want to check the validity of the whole string, then you need something like this:
^([^!]+!!)+$
^ start of the string
$ end of the string
It requires the whole string to contain only ([^!]+!!) one or more than one times.
If [^!] does not fit your requirements, you can of course replace it with [a-zA-Z] or similar.
I am new to using regex. I am trying to use the regex find and replace option in Notepad++.
I have used the following regex:
((?:)|(\+)|(-))(\d)((?:)|(\+)|(-))(/)((?:)|(\+)|(-))(\d)((?:)|(\+)|(-))
For the following text:
2/2
+2/+2
-2/-2
2+/2+
2-/2-
But I am able to get matches only for the first three. The last two, it only gives partial matches, excluding the last "+" and the "-". I am wondering if there is any upper limit for the number of groups (which i doubt is unlikely) that can be used or any upper limit for the maximum length of the regex. I am not sure why my regex is failing. Or if there is anything wrong with my regex, please correct it.
This is not an issue with Notepad++'s regex engine. The problem is that when you have alternations like (?:)|(\+)|(-), the regex engine will attempt to match the different options in the order they are specified. Since you specified an empty group first, it will attempt to match an empty string first, only matching the + or - if it needs to backtrack. This essentially makes the alternation lazy—it will never match any character unless it has to.
vks's answer works perfectly well, but just in case you actually needed those capturing groups separated out, you can do the same thing just by rewriting your alternations like this:
((\+)|(-)|(?:))(\d)((\+)|(-)|(?:))(/)((\+)|(-)|(?:))(\d)((\+)|(-)|(?:))
or even more simply, like this:
((\+)|(-)|)(\d)((\+)|(-)|)(/)((\+)|(-)|)(\d)((\+)|(-)|)
([-+]?)(\d)([-+]?)(/)([-+]?)(\d)([-+]?)
You can use this simple regex to match all cases.See here.
https://www.regex101.com/r/fG5pZ8/19
I am trying to create a regular expression that will capture several sections of a string. This is the expression I have created:
([0-9]{6}[-*][0-9xX]{7}).*([0-9]{1,3}-[0-9]{1,3}-[0-9]{1,3}).*([FPTSUCD])=?([01][*-])
The string that this runs against can appear in two different styles:
# 141803-6310114 #3-0-2 T0-jL
Or
]#0-7-4 C1-vU
When I use the first string I get all the parts I need.
141803-6310114
3-0-2
T
0-
When I use the second string I get no matches. This second sting is basically the same as the first but without this part “141803-6310114”. I would like the expression to work with both strings but for the number sequence to be optional. Can anyone advise on what the expression should look like to do this?
This will get you the parts in both cases:
(?:(\d{6}[-*][\dxX]{7}))?[^\d]*(\d{1,3}-\d{1,3}-\d{1,3}) ([FPTSUCD])=?([01][*-])
Made the first group optional (?) and changed the "eat all" between the first two groups to a "eat all non digits" + other clean up to make it more readable (at least to me ;)).
Regards