I've found some info on finding the first instance of a word in a string, but I'm trying to find the first instance of a word (two, actually, but in separate calls) only when it is preceded by some very specific text (an IP address delimited by underscores) that varies slightly. Also, these words are separated by underscores, so for some reason \b isn't working for me.
Here's some example strings to test against one line at a time. Only bolded words should be matched.
192_168_10_2_card02_port01_other_text_with_card_or_port
10_22_1_200_card4_port5_another_string_with_port_or_card
something_else_with_card_or_port_in_it
And in a second call, I'd like to match a different word in these strings.
192_168_10_2_card02_port01_other_text_with_card_or_port
10_22_1_200_card4_port5_another_string_with_port_or_card
something_else_with_card_or_port_in_it
My regex flavor is POSIX regex (for PostgreSQL 9.4). I've been able to run with anything that works in here http://regexpal.com/ so far.
Even if it can't solve for all 3 examples at once, if it could just solve for the first two, that would be very helpful.
Edit: To be absolutely clear, my intent is to replace the first string 'card' with the character 'c' and then to replace the first string 'port' with the letter 'p' without affecting any instance of 'card' or 'port' that are not immediately followed by numbers. This is why my match needs to include just those first words without their corresponding numbers.
If you can use negative lookahead you can use card((?!port).)*port to match a string with card, than any number of characters not followed by port, then card again.
EDIT:
if the input is always in the same format, then you can be more specific by using card[0-9]{1,2}_port. This will keep it from matching any other extraneous instances of card and port
EDIT2:
to match only the word in the first case you can use a positive lookahead: card(?=[0-9]{1,2}_port). Im not sure if your flavor allows positive lookbehind (the tester doesnt, but that is in js), but give (?<=card[0-9]{1,2}_)port a shot. If positive lookbehind doesnt work you may need to look into alternatives.
The \b assertion is not working in this case because _ is considered a word character.
Demo
You can use a look behind:
(?<=_)(card).*?(?<=_)(port)
Demo
To be even more specific, use the IP address pattern:
(^(?:\d+_){4})(card\d+)_(port\d+)
Demo
I had to solve this in two steps. In the first, I matched only lines with the IP string in the beginning (this excludes lines like my 3rd example). In the second step, I used regexp_replace to replace the first match of each word.
Unfortunately, I had completely missed the fact that regexp_replace only replaces the first match unless told otherwise with the 'g' flag:
WHEN (SELECT regexp_matches(mystring, '^1(?:[0-9]{1,3}_){4}card[0-9]{1,2}_port[0-9]{1,2}')) IS NOT NULL
THEN regexp_replace(regexp_replace(mystring, 'card', 'c'), 'port', 'p')
Though I still wish I could figure out how to match one of those words in a single expression, and I would accept any answer that could achieve that.
Related
I am in the process of learning Regex and have been stuck on this case. I have a url that can be in two states EXAMPLE 1:
spotify.com/track/1HYcYZCOpaLjg51qUg8ilA?si=Nf5w1q9MTKu3zG_CJ83RWA
OR EXAMPLE 2:
spotify.com/track/1HYcYZCOpaLjg51qUg8ilA
I need to extract the 1HYcYZCOpaLjg51qUg8ilA ID
So far I am using this: (?<=track\/)(.*)(?=\?)? which works well for Example 2 but it includes the ?si=Nf5w1q9MTKu3zG_CJ83RWA when matching with Example 1.
BUT if I remove the ? at the end of the expression then it works for Example 1 but not Example 2! Doesn't that mean that last group (?=\?) is optional and should match?
Where am I going wrong?
Thanks!
I searched a handful of "Questions that may already have your answer" suggestions from SO, and didn't find this case, so I hope asking this is okay!
The capturing group in your regular expression is trying to match anything (.) as much as possible due to the greediness of the quantifier (*).
When you use:
(?<=track\/)(.*)(?=\?)
only 1HYcYZCOpaLjg51qUg8ilA from the first example is captured, as there is no question mark in your second example.
When using:
(?<=track\/)(.*)(?=\??)
You are effectively making the positive lookahead optional, so the capturing group will try to match as much as possible (including the question mark), so that 1HYcYZCOpaLjg51qUg8ilA?si=Nf5w1q9MTKu3zG_CJ83RWA and 1HYcYZCOpaLjg51qUg8ilA are matched, which is not the desired output.
Rather than matching anything, it is perhaps more appropriate for you to match alphanumerical characters \w only.
(?<=track\/)(\w*)(?=\??)
Alternatively, if you are expecting other characters , let's say a hyphen - or a underscore _, you may use a character class.
(?<=track\/)([a-zA-Z0-9_-]*)(?=\??)
Or you might want to capture everything except a question mark ? with a negated character class.
(?<=track\/)([^?]*)(?=\??)
As pointed out by gaganso, a look-behind is not necessary in this situation (or indeed the lookahead), however it is indeed a good idea to start playing around with them. The look-around assertions do not actually consume the characters in the string. As you can see here, the full match for both matches only consists of what is captured by the capture group. You may find more information here.
This should work:
track\/(\w+)
Please see here.
Since track is part of both the strings, and the ID is formed from alphanumeric characters, the above regex which matches the string "track/" and captures the alphanumeric characters after that string, should provide the required ID.
Regex : (\w+(?=\?))|(\w+&)
See the demo for the regex, https://regexr.com/3s4gv .
This will first try to search for word which has '?' just after it and if thats unsuccessful it will fetch the last word.
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\/.
I need some help with building up my regex.
What I am trying to do is match a specific part of text with unpredictable parts in between the fixed words. An example is the sentence one gets when replying to an email:
On date at time person name has written:
The cursive parts are variable, might contains spaces or a new line might start from this point.
To get this, I built up my regex as such: On[\s\S]+?at[\s\S]+?person[\s\S]+?has written:
Basically, the [\s\S]+? is supposed to fill in any letter, number, space or break/new line as I am unable to predict what could be between the fixed words tha I am sure will always be there.
Now comes the hard part, when I would add the word "On" somewhere in the text above the sentence that I want to match, the regex now matches a much bigger text than I want. This is due to the use of [\s\S]+.
How am I able to make my regex match as less characters as possible? Using "?" before the "+" to make it lazy does not help.
Example is here with words "From - This - Point - Everything:". Cases are ignored.
Correct: https://regexr.com/3jdek.
Wrong because of added "From": https://regexr.com/3jdfc
The regex is to be used in VB.NET
A more real life, with html tags, can be found here. Here, I avoided using [\s\S]+? or (.+)?(\r)?(\n)?(.+?)
Correct: https://regexr.com/3jdd1
Wrong: https://regexr.com/3jdfu after adding certain parts of the regex in the text above. Although, in html, barely possible to occur as the user would never write the matching tag himself, I do want to make sure my regex is correctjust in case
These things are certain: I know with what the part of text starts, no matter where in respect to the entire text, I know with what the part of text ends, and there are specific fixed words that might make the regex more reliable, but they can be ommitted. Any text below the searched part is also allowed to be matched, but no text above may be matched at all
Another example where it goes wrong: https://regexr.com/3jdli. Basically, I have less to go with in this text, so the regex has less tokens to work with. Adding just the first < already makes the regex take too much.
From my own experience, most problems are avoided when making sure I do not use any [\s\S]+? before I did a (\r)?(\n)? first
[\s\S] matches all character because of union of two complementary sets, it is like . with special option /s (dot matches newlines). and regex are greedy by default so the largest match will be returned.
Following correct link, the token just after the shortest match must be geschreven, so another way to write without using lazy expansion, which is more flexible is to prepend the repeated chracter set by a negative lookahead inside loop,
so
<blockquote type="cite" [^>]+?>[^O]+?Op[^h]+?heeft(.+?(?=geschreven))geschreven:
becomes
<blockquote type="cite" [^>]+?>[^O]+?Op[^h]+?heeft((?:(?!geschreven).)+)geschreven:
(?: ) is for non capturing the group which just encapsulates the negative lookahead and the . (which can be replaced by [\s\S])
(?! ) inside is the negative lookahead which ensures current position before next character is not the beginning of end token.
Following comments it can be explicitly mentioned what should not appear in repeating sequence :
From(?:(?!this)[\s\S])+this(?:(?!point)[\s\S])+point(?:(?!everything)[\s\S])+everything:
or
From(?:(?!From|this)[\s\S])+this(?:(?!point)[\s\S])+point(?:(?!everything)[\s\S])+everything:
or
From(?:(?!From|this)[\s\S])+this(?:(?!this|point)[\s\S])+point(?:(?!everything)[\s\S])+everything:
to understand what the technic (?:(?!tokens)[\s\S])+ does.
in the first this can't appear between From and this
in the second From or this can't appear between From and this
in the third this or point can't appear between this and point
etc.
I am trying to use Regex to return the nth word in a string. This would be simple enough using other answers to similar questions; however, I do not have access to any of the code. I can only access a regex input field and the server only returns the 'full match' and cannot be made to return any captured groups such as 'group 1'
EDIT:
From the developers explaining the version of regex used:
"...its javascript regex so should mostly be compatible with perl i
believe but not as advanced, its fairly low level so wasn't really
intended for use by end users when originally implemented - i added
the dropdown with the intention of having some presets going
forwards."
/EDIT
Sample String:
One Two Three Four Five
Attempted solution (which is meant to get just the 2nd word):
^(?:\w+ ){1}(\S+)$
The result is:
One Two
I have also tried other variations of the regex:
(?:\w+ ){1}(\S+)$
^(?:\w+ ){1}(\S+)
But these just return the entire string.
I have tried replicating the behaviour that I see using regex101 but the results seem to be different, particularly when changing around the ^ and $.
For example, I get the same output on regex101 if I use the altered regex:
^(?:\w+ ){1}(\S+)
In any case, none of the comparing has helped me actually achieve my stated aim.
I am hoping that I have just missed something basic!
===EDIT===
Thanks to all of you who have contributed thus far, however, I am still running into issues. I am afraid that I do not know the language or restrictions on the regex other than what I can ascertain through trial and error, therefore here is a list of attempts and results all of which are trying to return "Two" from a sample of:
One Two Three Four Five
\w+(?=( \w+){1}$)
returns all words
^(\w+ ){1}\K(\w+)
returns no words atall (so I assume that \K does not work)
(\w+? ){1}\K(\w+?)(?= )
returns no words at all
\w+(?=\s\w+\s\w+\s\w+$)
returns all words
^(?:\w+\s){1}\K\w+
returns all words
====
With all of the above not working, I thought I would test out some others to see the limitations of the system
Attempting to return the last word:
\w+$
returns all words
This leads me to believe that something strange is going on with the start ^ and end $ characters, perhaps the server puts these in automatically if they are omitted? Any more ideas greatly appreciated.
I don't known if your language supports positive lookbehind, so using your example,
One Two Three Four Five
here is a solution which should work in every language :
\w+ match the first word
\w+$ match the last word
\w+(?=\s\w+$) match the 4th word
\w+(?=\s\w+\s\w+$) match the 3rd word
\w+(?=\s\w+\s\w+\s\w+$) match the 2nd word
So if a string contains 10 words :
The first and the last word are easy to find. To find a word at a position, then you simply have to use this rule :
\w+(?= followed by \s\w+ (10 - position) times followed by $)
Example
In this string :
One Two Three Four Five Six Seven Height Nine Ten
I want to find the 6th word.
10 - 6 = 4
\w+(?= followed by \s\w+ 4 times followed by $)
Our final regex is
\w+(?=\s\w+\s\w+\s\w+\s\w+$)
Demo
It's possible to use reset match (\K) to reset the position of the match and obtain the third word of a string as follows:
(\w+? ){2}\K(\w+?)(?= )
I'm not sure what language you're working in, so you may or may not have access to this feature.
I'm not sure if your language does support \K, but still sharing this anyway in case it does support:
^(?:\w+\s){3}\K\w+
to get the 4th word.
^ represents starting anchor
(?:\w+\s){3} is a non-capturing group that matches three words (ending with spaces)
\K is a match reset, so it resets the match and the previously matched characters aren't included
\w+ helps consume the nth word
Regex101 Demo
And similarly,
^(?:\w+\s){1}\K\w+ for the 2nd word
^(?:\w+\s){2}\K\w+ for the 3rd word
^(?:\w+\s){3}\K\w+ for the 4th word
and so on...
So, on the down side, you can't use look behind because that has to be a fixed width pattern, but the "full match" is just the last thing that "full matches", so you just need something whose last match is your word.
With Positive look-ahead, you can get the nth word from the right
\w+(?=( \w+){n}$)
If your server has extended regex, \K can "clear matched items", but most regex engines don't support this.
^(\w+ ){n}\K(\w+)
Unfortunately, Regex doesn't have a standard "match only n'th occurrence", So counting from the right is the best you can do. (Also, Regex101 has a searchable quick reference in the bottom right corner for looking up special characters, just remember that most of those characters are not supported by all regex engines)
I asked his question earlier but none of the responses solved the problem. Here is the full question:
Give a single UNIX pipeline that will create a file file1 containing all the words in file2, one word per line.Here a word is a string of letters, preceded and followed by nonletters.
I tried every single example that was given below, but i get "syntax error"s when using them.
Does anyone know how I can solve this??
Thanks
if your regex flavor support it you can use lookarounds:
(?<![a-zA-Z])[a-zA-Z]+(?![a-zA-Z])
(?<!..): not preceded by
(?!..): not followed by
If it is not the case you can use capturing groups and negated character classes:
(^|[^a-zA-Z])([a-zA-Z]+)($|[^a-zA-Z])
where the result is in group 2
^|[^a-zA-Z]: start of the string or a non letter characters (all character except letters)
$: end of the string
or the same with one capturing group and two non capturing groups:
(?:^|[^a-zA-Z])([a-zA-Z]+)(?:$|[^a-zA-Z])
(result in group 1)
In order to be unicode compatible, you could use:
(?:^|\PL)\pL+(?:\PL|$)
\pL stands for any letter in any language
\PL is the opposite of \pL
When your objective is to actually find words, the most natural way would be
\b[A-Za-z]+\b
However, this assumes normal word boundaries, like whitespaces, certain punctuations or terminal positions. Your requirement suggests you want to count things like the "example" in "1example2".
In that case, I would suggest using
[A-Za-z]+
Note that you don't actually need to look for what precedes or follows the alphabets. This already captures all alphabets and only alphabets. The greedy requirement (+) ensures that nothing is left out from a capture.
Lookarounds etc should not be necessary because what you want to capture and what you want to exclude are exact inverses of each other.
[Edit: Given the new information in comments]
The methods below are similar to Casimir's, except that we exclude words at terminals (which we were explicitly trying to capture, because of your original description).
Lookarounds
(?<=[^A-Za-z])[A-Za-z]+(?=[^A-Za-z])
Test here. Note that this uses negated positive lookarounds, and not Negative lookarounds as they would end up matching at the string terminals (which are, to the regex engine as much as to me, non-alphabets).
If lookarounds don't work for you, you'd need capturing groups.
Search as below, then take the first captured group.
[^A-Za-z]([A-Za-z]+)[^A-Za-z]
When talking about regex, you need to be extremely specific and accurate in your requirements.