I am trying to figure out how to look through a string and replace any word which ends at ".com" at the end with a valid url link.
for example:
"google.com has launces a .." will be replaced with
"<.a href='google.com'>google.com <./a> has launches .."
// I tried following code, but it only works for finding word which starts with "www."
data.rows[j].content = data.rows[j].content.replace(
/(^|<|\s)(www\..+?\..+?)(\s|>|$)/g,'$2'
)
First, I recommend learning how regex works. It's a very powerful tool that all developers should understand because it appears in many different programming languages.
Once you understand the basics, this should make more sense:
/(^|<|\b)(\S+?\.com)(\b|>|$)/g
Regex101 Demo - (for a breakdown of the regex, look in the top-right pane)
https://regex101.com/r/oF0mA9/2 should do the trick from the regex front.
You can follow this ways for ending com
(http(s?):)|([/|.|\w|\s])*\.(?:com)
OR
(?i)\.(com)$
OR
(?:([^:/?#]+):)?(?://([^/?#]*))?([^?#]*\.(?:com))(?:\?([^#]*))?(?:#(.*))?
Resource Link:
Regex to check if valid URL that ends in .jpg, .png, or .gif
Related
I'm trying to write a regex for use in Calibre (python) to find ebooks that have the series name in brackets in the title. I have a custom column with the series name and title separated by a "~", for example:
"The Series~The Book Title (The Series)"
Best I can come up with finds anything with at least one letter from the series name in brackets in the title:
(.+)~.*[\(\1\)].*
I only want to find those that have the whole of the first part of the string in brackets at the end of the second part, it can contain extra info.
Thanks.
This works in Notepad++:
(.+)~[^\(]*\(\1\).*
I'm not sure it will work the same in python, but regexp processors are usually very similar, so try it out.
Your regex is pretty close, you can change a little your regex and have this:
(.+?)~.*[([]\1[)\]].*
Working demo
This will match strings like:
The Series~The Book Title (The Series)
The Series~The Book Title [The Series]
However, if you just want to match words with paretheses, then you can have:
(.+?)~.*[(]\1[)].*
or
(.+?)~.*\(\1\).*
Working demo
Thanks for the suggestions. They work perfectly in the python demo but for some unknown reason don't work in Calibre. Seems like one character is the most it will match from the capture group. Must be a limitation in the regex system Calibre uses.
I created a regex string that I hoped would get both the link and the associated text in an html page. For instance, if I had a link such as:
<a href='www.la.com/magic.htm'>magicians of los angeles</a>
Then the link I want is 'www.la.com/magic.htm' and the text I want is 'magicians of los angeles'.
I used the following regex expression:
strsearch = "\<a\s+(.*?)\>(.*?)\</a\s*?\>|"
But my vb program told me I was getting too many matches.
Is there something wrong with the regEx expression?
The circle-brackets are meant to get 'groups' that can be back-referenced.
Thanks
What about this one:
\<a href=.+\</a>
All there is left to do is to go over each match and extract the substrings using regular string manipulation.
Check here (although regexr follows javascript regex implementation, it is still useful in our scenario)
With that being said, I often see people stating that regexes are not suited for parsing Html. You might need to use an Html Parser for this. You have HtmlAgilityPack, which is not maintained anymore, and AngleSharp, that I know of to recommend.
I tried with following pattern , it worked.
\<a href=(.*?)\>(.*?)\<\/a\s*?\>|
Also Found two errors on your origin string:
missed a escape syntax on /a
the reserved word 'href' is captured on
first group
At last , i would like recommend you a great site to test REGEX string. It will helps your debug really fast. Refer this (also demonstrating the result you want) :
REGEX101
I'm currently trying to match all plain text links in a markdown text.
Example of the markdown text:
Dude, look at this url http://www.google.com .. it's a great search engine
I would like it to be converted into
Dude, look at this url <http://www.google.com> .. it's a great search engine
So in short, processing url should become <url>, but processing existing <url> shouldnt become <<url>>. Also, the link in the markdown can be in the form of (url), so we'll have to avoid matching the normal brackets too.
So my working regex for matching the plain text url in java is :
"[^(\\<|\\(](https?|ftp|file)://[-a-zA-Z0-9+&##/%?=~_|!:,.;]*[-a-zA-Z0-9+&##/%=~_|][^(\\>|\\)]",
with [^(\\<|\\(] and [^(\\>|\\)] to avoid matching the wrapping brackets.
But here lies one problem where i also do not want to match this kind of url :
[1]: http://slashdot.org
So, if the markdown text is
Dude, look at this url http://www.google.com .. it's a great search engine
[1]: http://slashdot.org
I want only http://www.google.com to be matched, but not the http://slashdot.org.
I wonder what's the pattern to meet this criteria ?
What you have here is a parsing problem. Regexes are fine, but just using regexes here will make it a mess (supposing you achieve it). After you fix this problem, you'll probably find yourself facing other ones, like URL in code (between ` or in lines starting with tabs or four spaces) that you don't want to replace.
A solution would be to split into lines and then
detect patterns (for example ^\[\d+\]:\s+)
apply your replacements (for example this URL to link change) only on lines which doesn't follow an incompatible pattern
That's the logic I use in this small pseudo-markdown parser that you can test here.
Note that there's always the solution to use an existing proved markdown parser, there are many of them.
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 :)
I am trying to contruct a regular expression to remove links from content unless it contains 1 of 2 conditions.
<a.*?href=[""'](http[s]?:\/\/(.*?)\.link\.com)?\/(?!m\/).*?<\/a>
This will match any link to link.com that does not have m/ at the end of the domain section. I want to change this slightly so it does't match URLs that are links to pdf files regardless of having the m/ in the url, I came up with:
<a.*?href=["'](http[s]?:\/\/(.*?)\.brodies\.com)?\/(?!m\/).*?\.(?!pdf)["'].*?<\/a>
Which is ooh so very close except now it will only match if the URL has a "." at the end - I can see why it's doing it. I can't seem to make the "." optional as this causes the non greedy pattern prior to the "." to keep going until it hits the ["']
Any help would be good to help solve this.
Thanks
Paul
You probably want to use (?<!\.pdf)["'] instead of \.(?!pdf)["'].
But note that this expression has several issues, best way to solve them is to use a proper HTML parser.
First, RegEx match open tags except XHTML self-contained tags.
That said, (since it probably will not deter,) here is a slightly-better-constrained version of what you're trying to, with the caveat that this is still not good enough!
<a[^>]+?href\s*=\s*["'](https?:\/\/[^"']*?\.link\.com)?\/(?!m\/)[^"']*?\.(?!pdf)[^"']*?["'][^>]*?>.*?<\/a>
You can see a running example of this regex at: http://rubular.com/r/obkKrKpB8B.
Your problem was actually just that you were looking for a quote character immediately after the dot, here: .(?!pdf)["'].