I'm working on a simple router, and I need to be able to identify if there's a query-like structure at the end of a URL address.
What I came up with:
(\?([^&=]+)=([^&=]+)&?)+$
simply does not work! It works on a first iteration: i.e. xxx?foo=bar, but definitely not on two i.e. xxx?foo=bar&greeting=hello won't work.
What am I doing wrong? And also: Is there a better solution to accomplish this?
You need to match one key-value pair (([^&=]+)=([^&=]+)) preceded with a question mark (\?([^&=]+)=([^&=]+)) followed by zero to any number of key-value pairs preceded by an ampersand each ((?:&([^&=]+)=([^&=]+))*):
\?([^&=]+)=([^&=]+)(?:&([^&=]+)=([^&=]+))*$
Demo: https://regex101.com/r/OHdRHS/1
Related
As you know, Google links can be pretty unwieldy:
https://www.google.com/search?q=some+search+here&source=hp&newwindow=1&ei=A_23ssOllsUx&oq=some+se....
I have MANY Google links saved that I would like to clean up to make them look like so:
https://www.google.com/search?q=some+search+here
The only issue is that I cannot figure out the correct regex pattern for Vim to do this.
I figure it must be something like this:
:%s/&source=[^&].*//
:%s/&source=[^&].*[^&]//
:%s/&source=.*[^&]//
But none of these are working; they start at &source, and replace until the end of the line.
Also, the search?q=some+search+here can appear anywhere after the .com/, so I cannot rely on it being in the same place every time.
So, what is the correct Vim regex pattern to use in order to clean up these links?
Your example can easily be dealt with by using a very simple pattern:
:%s/&.*
because you want to keep everything that comes before the second parameter, which is marked by the first & in the string.
But, if the q parameter can be anywhere in the query string, as in:
https://www.google.com/search?source=hp&newwindow=1&q=some+search+here&ei=A_23ssOllsUx&oq=some+se....
then no amount of capturing or whatnot will be enough to cover every possible case with a single pattern, let alone a readable one. At this point, scripting is really the only reasonable approach, preferably with a language that understands URLs.
--- EDIT ---
Hmm, scratch that. The following seems to work across the board:
:%s#^\(https://www.google.com/search?\)\(.*\)\(q=.\{-}\)&.*#\1\3
We use # as separator because of the many / in a typical URL.
We capture a first group, up to and including the ? that marks the beginning of the query string.
We match whatever comes between the ? and the first occurrence of q= without capturing it.
We capture a second group, the q parameter, up to and excluding the next &.
We replace the whole thing with the first capture group followed by the second capture group.
I'm trying to get regexp code for the below case. I tried multiple tries but in vain.
I need to catch any URLs of the domain site.com. Tried using regexp '^site.com/*$
but it does not recognizes it.
i'm just looking for regexp code whichmatches site.com/*
With your expression ^site.com/*$ you match all strings that start with site.com and have zero or more trailing / characters (/*):
If you want to match any strings starting with site.com/ you might want to try ^site\.com/.*$:
There are already a lot of other regex questions regarding domain names on SO, but your question is not clear to me in what context you are trying to do this, or what is the actual goal you want to achieve. If you describe your needs more precisely you could probably find some answers on this forum.
I generally use a helper website like regex101.com.
Also, a few things to note, . has a special meaning in regex meaning any character, and if you wanted to capture site.com/foo you might want to use something where you are not limited to the number of characters by the end. I'd do this with groupings.
^(site\.com\/)(.+)$
You can see this in action here: https://regex101.com/r/AU2iYC/2
Your regex ^site.com/*$ is only matched follow sentences
ex) site.com/ site.com//////// site.com
because * asterisk in regex means Match 0 or more of the preceding token.
so, it should be work
^site.com\/.*$
Given this string: hello"C07","73" (quotes included) I want to get "C07". I'm using (?:hello)|(?<=")(?<screen>[a-zA-Z0-9]+)?(?=") to try to do this. However, it consistently matches "73" as well. I've tried ...0-9]+){1}..., but that doesn't work either. I must be misunderstanding how this is supposed to work, but I can't figure out any other way.
How can I get just the first set of characters between quotes?
EDIT: Here's a link to show my problem.
EDIT: Ok, here's exactly what I'm trying to do:
Basically, what I'm trying to get is this: 1) a positive match on "hello", 2) a group named "screen" with, in this case, "C07" in it and 3) a group named "format" with, in this case, "73" in it.
Both the "C07" and "73" will vary. "hello" will always be the same. There may or may not be an extra comma between "hello" and the first double-quote.
For you initial question of how to stop after the first match either removing the global search, or searching from the start of the string would accomplish that.
For the latter question you can name your groups and just keep extending the pattern throughout the line(s).
hello"(?<screen>[^"]+)","(?<format>[^"]+)"
Demo: http://regex101.com/r/PBXe8l/1
Based on your regex example, why not:
^(?:hello)"([a-zA-Z\d]+)"
Regex Demo
Given an XML document, I'd like to be able to pick out individual key/value pairsfrom a particular tag:
<aaa>key0:val0 key1:val1 key2:va2</aaa>
I'd like to get back
key0:val0
key1:val1
key2:val2
So far I have
(?<=<aaa>).*(?=<\/aaa>)
Which will match everything inside, but as one result.
I also have
[^\s][\w]*:[\w]*[^\s] which will also match correctly in groups on this:
key0:val0 key1:val1 key2:va2
But not with the tags. I believe this is an issue with searching for subgroups and I'm not sure how to get around it.
Thanks!
You cannot combine the two expressions in the way you want, because you have to match each occurrence of "key:value".
So in what you came up with - (?<=<abc>)([\w]*:[\w]*[\s]*)+(?=<\/abc>) - there are two matching groups. The bigger one matches everything inside the tags, while the other matches a single "key:value" occurrence. The regex engine cannot give each individual occurence because it does not work that way. So it just gives you the last one.
If you think in python, on the matcher object obtained after applying you regex, you will have access to matcher.group(1) and matcher.group(2), because you have two matching ( ) groups in the regex.
But what you want is the n occurences of "key:value". So it's easier to just run the simpler \w+:\w+ regex on the string inside the tags.
I uploaded this one at parsemarket, and I'm not sure its what you are looking for, but maybe something like this:
(<aaa>)((\w+:\w+\s)*(\w+:\w+)*)(<\/aaa>)
AFAIK, unless you know how many k:v pairs are in the tags, you can't capture all of them in one regex. So, if there are only three, you could do something like this:
<(?:aaa)>(\w+:\w+\s*)+(\w+:\w+\s*)+(\w+:\w+\s*)+<(?:\/aaa)>
But I would think you would want to do some sort of loop with whatever language you are using. Or, as some of the comments suggest, use the parser classes in the language. I've used BeautifulSoup in Python for HTML.
I found this simple regexp (i know it's probably not perfect) somewhere online to validate an email address.
/^(?:\w+\.?)*\w+#(?:\w+\.)+\w+$/
The problem is, that this regexp doesn't allow for the following case:
myname#test-domain.com
my-name#test-domain.com
Any ideas?
ps. I'm using this regexp within javascript.
If you simply want to add hyphens you can change the regexp to:
/^(?:\w+[\-\.])*\w+#(?:\w+[\-\.])*\w+\.\w+$/
To add other special chars e.g. like underscore just put them in the first (not the second) pair of square brackets, i.e. change [\-\.] to [\-\._].
Also have a look on this question and its anwer.