Let's take an url like
www.url.com/some_thing/random_numbers_letters_everything_possible/set_of_random_characters_everything_possible.randomextension
If I want to capture "set_of_random_characters_everything_possible.randomextension" will [^/\n]+$work? (solution taken from Trying to get the last part of a URL with Regex)
My question is: what does the "\n" part mean (it works even without it)? And, is it secure if the url has the most casual combination of characters apart "/"?
First, please note that www.url.com/some_thing/random_numbers_letters_everything_possible/set_of_random_characters_everything_possible.randomextension is not a URL without a scheme like http:// in front of it.
Second, don't parse URLs yourself. What language are you using? You probably don't want to use a regex, but rather an existing module that has already been written, tested, and debugged.
If you're using PHP, you want the parse_url function.
If you're using Perl, you want the URI module.
Have a look at this explanation: http://regex101.com/r/jG2jN7
Basically what is going on here is "match any character besides slash and new line, infinite to 1 times". People insert \r\n into negated char classes because in some programs a negated character class will match anything besides what has been inserted into it. So [^/] would in that case match new lines.
For example, if there was a line break in your text, you would not get the data after the linebreak.
This is however not true in your case. You need to use the s-flag (PCRE_DOTALL) for this behavior.
TL;DR: You can leave it or remove it, it wont matter.
Ask away if anything is unclear or I've explained it a little sloppy.
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 am currently learning regex and I am trying to filter all links (eg: http://www.link.com/folder/file.html) from a document with notepad++. Actually I want to delete everything else so that in the end only the http links are listed.
So far I tried this : http\:\/\/www\.[a-zA-Z0-9\.\/\-]+
This gives me all links which is find, but how do I delete the remaining stuff so that in the end I have a neat list of all links?
If I try to replace it with nothing followed by \1, obviously the link will be deleted, but I want the exact opposite to have everything else deleted.
So it should be something like:
- find a string of numbers, letters and special signs until "http"
- delete what you found
- and keep searching for more numbers, letters ans special signs after "html"
- and delete that again
Any ideas? Thanks so much.
In Notepad++, in the Replace menu (CTRL+H) you can do the following:
Find: .*?(http\:\/\/www\.[a-zA-Z0-9\.\/\-]+)
Replace: $1\n
Options: check the Regular expression and the . matches newline
This will return you with a list of all your links. There are two issues though:
The regex you provided for matching URLs is far from being generic enough to match any URL. If it is working in your case, that's fine, else check this question.
It will leave the text after the last matched URL intact. You have to delete it manually.
The answer made previously by #psxls was a great help for me when I have wanted to perform a similar process.
However, this regex rule was written six years ago now: accordingly, I had to adjust / complete / update it in order it can properly work with the some recent links, because:
a lot of URL are now using HTTPS instead of HTTP protocol
many websites less use www as main subdomain
some links adds punctuation mark (which have to be preserved)
I finally reshuffle the search rule to .*?(https?\:\/\/[a-zA-Z0-9[:punct:]]+) and it worked correctly with the file I had.
Unfortunately, this seemingly simple task is going to be almost impossible to do in notepad++. The regex you would have to construct would be...horrible. It might not even be possible, but if it is, it's not worth it. I pretty much guarantee that.
However, all is not lost. There are other tools more suitable to this problem.
Really what you want is a tool that can search through an input file and print out a list of regex matches. The UNIX utility "grep" will do just that. Don't be scared off because it's a UNIX utility: you can get it for Windows:
http://gnuwin32.sourceforge.net/packages/grep.htm
The grep command line you'll want to use is this:
grep -o 'http:\/\/www.[a-zA-Z0-9./-]\+\?' <filename(s)>
(Where <filename(s)> are the name(s) of the files you want to search for URLs in.)
You might want to shake up your regex a little bit, too. The problems I see with that regex are that it doesn't handle URLs without the 'www' subdomain, and it won't handle secure links (which start with https). Maybe that's what you want, but if not, I would modify it thusly:
grep -o 'https\?:\/\/[a-zA-Z0-9./-]\+\?' <filename(s)>
Here are some things to note about these expressions:
Inside a character group, there's no need to quote metacharacters except for [ and (sometimes) -. I say sometimes because if you put the dash at the end, as I have above, it's no longer interpreted as a range operator.
The grep utility's syntax, annoyingly, is different than most regex implementations in that most of the metacharacters we're familiar with (?, +, etc.) must be escaped to be used, not the other way around. Which is why you see backslashes before the ? and + characters above.
Lastly, the repetition metacharacter in this expression (+) is greedy by default, which could cause problems. I made it lazy by appending a ? to it. The way you have your URL match formulated, it probably wouldn't have caused problems, but if you change your match to, say [^ ] instead of [a-zA-Z0-9./-], you would see URLs on the same line getting combined together.
I did this a different way.
Find everything up to the first/next (https or http) (then everything that comes next) up to (html or htm), then output just the '(https or http)(everything next) then (html or htm)' with a line feed/ carriage return after each.
So:
Find: .*?(https:|http:)(.*?)(html|htm)
Replace with: \1\2\3\r\n
Saves looking for all possible (incl non-generic) url matches.
You will need to manually remove any text after the last matched URL.
Can also be used to create url links:
Find: .*?(https:|http:)(.*?)(html|htm)
Replace: \1\2\3\r\n
or image links (jpg/jpeg/gif):
Find: .*?(https:|http:)(.*?)(jpeg|jpg|gif)
Replace: <img src="\1\2\3">\r\n
I know my answer won't be RegEx related, but here is another efficient way to get lines containing URLs.
This won't remove text around links like Toto mentioned in comments.
At least if there is nice pattern to all links, like https://.
CTRL+F => change tab to Mark
Insert https://
Tick Mark to bookmark.
Mark All.
Find => Bookmarks => Delete all lines without bookmark.
I hope someone who lands here in search of same problem will find my way more user-friendly.
You can still use RegEx to mark lines :)
Using regex I'm trying to remove empty bbcode tags. By empty I mean nothing in between them:
[tag][/tag]
If there is something between them then it should be kept.
I've searched a lot and played around with a regex tester but haven't come up with anything that works right.
Edit: I realize now why I was having a hard time with this. In addition to the example above, I also have one's like:
[url=http://www.somedomain.com/][/url]
I'm trying to cleanup bbcode when a form is submitted so it's not stored since it's unneeded.
In Javascript, you could do :
str.replace(/\[([^\[\]]*)\]\[\/\1\]/g, '');
The operative aspect of regex in this case is the use of internal backrefs; I'm not sure, off the top of my head, whether this is universally supported, but .NET, in any case, seems to use PCRE (is this true?).
The pattern, then, is [, a word, ][/, the same word, ]. If we assume the word has simply the quality of "does not contain ]", then an appropriate regex to match an empty tag is \[([^\]]+)\]\[/\1\], escaped as necessary in context.
For the second case, if assume the form [tag=arg][/tag], and that tag and arg each don't contain any ']' (not a reasonable assumption! but dealing with it is left as an exercise for the reader -- and I'm quite sure most bbcode implementations don't actually deal with that problem, either), one could use a regex \[([^\]=]+)(=[^\]]*)?\]\[/\1\].
We're using a proprietary tracking system that requires the use of regular expressions to load third party scripts on the URLs we specify.
I wanted to check the syntax of the regex we're using to see if it looks right.
To match the following URL
/products/18/indoor-posters
We are using this rule:
.*\/products\/18\/indoor-posters.*
Does this look right? Also, if there was a query parameter on the URL, would it still work? e.g.
/products/18/indoor-posters?someParam=someValue
There's another URL to match:
/products
The rule for this is:
.*\/products
Would this match correctly?
Well, "right" is a relative term. Usually, .* is not a good idea because it matches anything, even nothing. So while these regexes will all match your example strings, they'll also match much more. The question is: What are you using the regexes for?
If you only want to check whether those substrings are present anywhere in the string, then they are fine (but then you don't need regex anyway, just check for substrings).
If you want to somehow check whether it's a valid URL, then no, the regexes are not fine because they'd also match foo-bar!$%(§$§$/products/18/indoor-postersssssss)(/$%/§($/.
If you can be sure that you'll always get a correct URL as your input and just want to check whether they match you pattern, then I'd suggest
^.*\/products$
to match any URL that ends in /products, and
^.*\/products\/18\/indoor-posters(?:\?[\w-]+=[\w-]+)?$
to match a URL that ends in /products/18/indoor-posters with an optional ?name=value bit at the end, assuming only alphanumeric characters are legal for name and value.
I am working with legacy systems at the moment, and a lot of work involves breaking up delimited strings and testing against certain rules.
With this string, how could I return "Active" in a back reference and search terms, stopping when it hits the first caret (^)?:
Active^20080505^900^LT^100
Can it be done with an inclusion in the regex of this "(.+)" ? The reason I ask is that the actual regex "(.+)" is defined in a database as cutting up these messages and their associated rules can be set from a front-end system. The content could be anything ('Active' in this case), that's why ".+" has been used in this case.
Rule: The caret sign cannot feature between the brackets, as that would result with it being stored in the database field too, and it is defined elsewhere in another system field.
If you have a better suggestion than "(.+)" will be happy to hear it.
Thanks in advance.
(.+?)\^
Should grab up to the first ^
If you have to include (.+) w/o modifications you could use this:
(.+?)\^(.+)
The first backreference will still be the correct one and you can ignore the second.
A regex is really overkill here.
Just take the first n characters of the string where n is the position of the first caret.
Pseudo code:
InputString.Left(InputString.IndexOf("^"))
^([^\^]+)
That should work if your RE library doesn't support non-greediness.