Regex lookahead with multiple negative conditions - regex

I am performing a regex on a HTML string to fetch URL's. I want to fetch all href's and src's that are not javascript. From another SO post I have the following pattern:
/(href|src)?\="http:\/\/www\.mydomain\.com\/(?:(?!\.js).)*"/
Which fetches me results like:
src="http://www.mydomain.com/path/to/resource/image.gif" alt="" border="0"
This is good because it is missing the .js results. It's bad because it's fetching additional tags in the element. I tried the following amendment to stop at the first ":
/(href|src)?\="http:\/\/www\.mydomain\.com\/(?:(?!\.js).)[^"]*"/
It works in that it returns href="$url", but it returns results ending in .js. Is there a way to combine a negative lookahead that says:
Match string until it comes across another " - i.e. [^"]*; and
Do not match string if it ends in .js"
Thanks in advance for any help/tips/pointers.

add a "?" to the "*" before the last quote. This will make the "*" non-greedy, ie: it will stop matching at the first quote, not the last
/(href|src)?\="http:\/\/www\.mydomain\.com\/(?:(?!\.js).)*?"/

Here's something a bit different. I used Debuggex with this expression:
(?:src|href)=(?&.quotStr)(?<!\.js")
which compiled it to this one:
$regex = '/(?:src|href)=(?:"((?:\\\\.|[^"\\\\]){0,})")(?<!\\.js")/';
Live Demo

If you only want to reject .js at the end of the string, you can use the following for the last part of the string match:
"(?![^"]*\.js").*?"
per this Rubular

EDIT
See: https://stackoverflow.com/a/18838123/1163653 for a better solution.
Fixed it:
/(href|src)?\="http:\/\/www\.mydomain\.com\/(?:(?!\.js"|").)*"/
Note that the lookahead is checking for any string (after the domain) that doesn't contain .js or ", both of which would cause it to be invalid. It allows hrefs ending in .css through as they only fail when they reach the first ", which is the behaviour needed.

Related

How to extract file name from URL?

I have file names in a URL and want to strip out the preceding URL and filepath as well as the version that appears after the ?
Sample URL
Trying to use RegEx to pull, CaptialForecasting_Datasheet.pdf
The REGEXP_EXTRACT in Google Data Studio seems unique. Tried the suggestion but kept getting "could not parse" error. I was able to strip out the first part of the url with the following. Event Label is where I store URL of downloaded PDF.
The URL:
https://www.dudesolutions.com/Portals/0/Documents/HC_Brochure_Digital.pdf?ver=2018-03-18-110927-033
REGEXP_EXTRACT( Event Label , 'Documents/([^&]+)' )
The result:
HC_Brochure_Digital.pdf?ver=2018-03-18-110927-033
Now trying to determine how do I pull out everything after the? where the version data is, so as to extract just the Filename.pdf.
You could try:
[^\/]+(?=\?[^\/]*$)
This will match CaptialForecasting_Datasheet.pdf even if there is a question mark in the path. For example, the regex will succeed in both of these cases:
https://www.dudesolutions.com/somepath/CaptialForecasting_Datasheet.pdf?ver
https://www.dudesolutions.com/somepath?/CaptialForecasting_Datasheet.pdf?ver
Assuming that the name appears right after the last / and ends with the ?, the regular expression below will leave the name in group 1 where you can get it with \1 or whatever the tool that you are using supports.
.*\/(.*)\?
It basically says: get everything in between the last / and the first ? after, and put it in group 1.
Another regular expression that only matches the file name that you want but is more complex is:
(?<=\/)[^\/]*(?=\?)
It matches all non-/ characters, [^\/], immediately preceded by /, (?<=\/) and immediately followed by ?, (?=\?). The first parentheses is a positive lookbehind, and the second expression in parentheses is a positive lookahead.
This REGEXP_EXTRACT formula captures the characters a-zA-Z0-9_. between / and ?
REGEXP_EXTRACT(Event Label, "/([\\w\\.]+)\\?")
Google Data Studio Report to demonstrate.
Please try the following regex
[A-Za-z\_]*.pdf
I have tried it online at https://regexr.com/. Attaching the screenshot for reference
Please note that this only works for .pdf files
Following regex will extract file name with .pdf extension
(?:[^\/][\d\w\.]+)(?<=(?:.pdf))
You can add more extensions like this,
(?:[^\/][\d\w\.]+)(?<=(?:.pdf)|(?:.jpg))
Demo

How to extract FirstName and LastName from html tags with regex?

I have response body which contains
"<h3 class="panel-title">Welcome
First Last </h3>"
I want to fetch 'First Last' as a output
The regular expression I have tried are
"Welcome(\s*([A-Za-z]+))(\s*([A-Za-z]+))"
"Welcome \s*([A-Za-z]+)\s*([A-Za-z]+)"
But not able to get the result. If I remove the newline and take it as
"<h3 class="panel-title">Welcome First Last </h3>" it is detecting in online regex maker.
I suspect your problem is the carriage return between "Welcome" and the user name. If you use the "single-line mode" flag (?s) in your regex, it will ignore newlines. Try these:
(?s)Welcome(\s*([A-Za-z]+))(\s*([A-Za-z]+))
(?s)Welcome \s*([A-Za-z]+)\s*([A-Za-z]+)
(this works in jMeter and any other java or php based regex, but not in javascript. In the comments on the question you say you're using javascript and also jMeter - if it is a jMeter question, then this will help. if javaScript, try one of the other answers)
Well, usually I don't recommend regex for this kind of work. DOM manipulation plays at its best.
but you can use following regex to yank text:
/(?:<h3.*?>)([^<]+)(?:<\/h3>)/i
See demo at https://regex101.com/r/wA2sZ9/1
This will extract First and Last names including extra spacing. I'm sure you can easily deal with spaces.
In jmeter reg exp extractor you can use:
<h3 class="panel-title">Welcome(.*?)</h3>
Then take value using $1$.
In the data you shown welcome is followed by enter.If actually its part of response then you have to use \n.
<h3 class="panel-title">Welcome\n(.*?)</h3>
Otherwise above one is enough.
First verify this in jmeter using regular expression tester of response body.
Welcome([\s\S]+?)<
Try this, it will definitely work.
Regular expressions are greedy by default, try this
Welcome\s*([A-Za-z]+)\s*([A-Za-z]+)
Groups 1 and 2 contain your data
Check it here

Exclude a slash in capture group of regex

i have string, which has the value like
"urls":[
{
"url":"https:\/\/t.co\/OjiDUThEvK",
"expanded_url":"http:(escape sequence slash)/(escape sequence slash)/fb.me\/7Wnh0hMLL",
"display_url":"fb.me(escape sequence slash)/7Wnh0hMLL",
"indices":[48,71]}],
"user_mentions":[],
"symbols":[]
}
]
i need to capture only "expanded url" i tried the following regex:
"expanded_url"\:\"http\:\\\/\\\/(.*?)\"
this gave a result :
"fb.me(escape sequence slash)/7Wnh0hMLL"
but i want to exclude the escape sequence slash in the URL, is it possible to achieve the same, kindly let me know the changes to me made to the regex
I'm not 100% sure if this is what you're after. Can you post the raw input without the "(escape sequence slash)" part I'm assuming that this is actually / in the text you're matching against.
match:
\"expanded_url\":\"http:\\\/\\\/([^\\]*)\\\/([^\\"]*)\"
replace with:
$1/$2

Regex: matching unknown groups that repeat?

I'm trying to create a generic regex pattern for a crawler, to avoid so called "crawler traps" (links that just add url parameters and refer to the exact same page, which results in tons of useless data). Alot of times, those links just add the same part to a URL over and over again. Here is an example out of a log file:
http://examplepage.com/cssms/chrome/cssms/chrome/cssms/pages/browse/cssms/pages/misc/...
I can use regular expressions to narrow the scope of the crawler and i would love to have a pattern, that tells the crawler to ignore everything that has repeating parts. Is that possible with a regex?
Thanks in advance for some tips!
JUST TO CLARIFY:
the crawlertraps are not designed to prevent crawling, they are a result of poor web design. All the pages we are crawling explicitly allowed us to do so!
If you are already looping through a list of URLs, you could add matching as a condition to skip the current iteration:
array = ["/abcd/abcd/abcd/abcd/", "http://examplepage.com/cssms/chrome/cssms/chrome/cssms/pages/browse/cssms/pages/misc/", "http://examplepage/apple/cake/banana/"]
import re
pattern1 = re.compile(r'.*?([^\/\&?]{4,})(?:[\/\&\?])(.*?\1){3,}.*')
for url in array:
if re.match(pattern1, url):
print "It matches; skipping this URL"
continue
print url
Example regex:
.*?([^\/\&?]{4,})(?:[\/\&\?])(.*?\1){3,}.*
([^\/\&?]{4,}) matches and captures sequences of anything, but not containing [/&?], repeated 4 or more times.
(?:[\/\&\?]) looks for one /,& or ?
(.*?(?:[\/\&\?])\1){3,} match anything until [/&?], followed by what we captured, doing all of this 3 or more times.
demo
You can use a backreference in Python/PERL regexes (and possibly others) to catch a pattern which is repeated:
>>> re.search(r"(/.+)\1", "http://examplepage.com/cssms/chrome/cssms/chrome/cssms/pages/browse/cssms/pages/misc/").group(1)
'/cssms/chrome'
\1 references the first match, so (/.+)\1 means the same sequence repeated twice in a row. The leading / is just to avoid the regex matching the first single repeating letter (which is the t in http) and catch repetitions in the path.

MFC: How do I construct a good regular expression that validates URLs?

Here's the regular expression I use, and I parse it using CAtlRegExp of MFC :
(((h|H?)(t|T?)(t|T?)(p|P?)(s|S?))://)?([a-zA-Z0-9]+[\.]+[a-zA-Z0-9]+[\.]+[a-zA-Z0-9])
It works fine except with one flaw. When URL is preceded by characters, it still accepts it as a URL.
ex inputs:
this is a link www.google.com (where I can just tokenize the spaces and validate each word)
is...www.google.com (this string still matches the RegEx above :( )
Please help...
Thanks...
Use the IgnoreCase flag instead of catering for each case.
Stick a ^ at the beginning if you want the start of the string to be the start of the URL
You're missing a lot of characters from possible, valid URLs.
You need to tell the regex to only match at the start and end of the string. I'm not sure how you do that in VC++ - in most regexs you enclose the pattern with ^ and $. The ^ says "the start of the string" and the $ says "the end of the string."
^(((h|H?)(t|T?)(t|T?)(p|P?)(s|S?))\://)?([a-zA-Z0-9]+[\\.]+[a-zA-Z0-9]+[\\.]+[a-zA-Z0-9])$
The second is matching because the string still contains a valid URL.
How about using CUrl (that is, 'C-Url', in ATL, not curl as in libcurl) which can 'parse' urls with CUrl::CrackUrl . If that function returns FALSE you assume it's not a valid URL.
That said, decomposing URL is sufficiently complex to warrant a proper parser, not a regex based decomposition. Cfr. rfc 2396 etc. for an overview on the complexities.
Start the regex with ^ to and end it with $ to have the regex match only if the entire sting matches (if that's what you want):
^(((h|H?)(t|T?)(t|T?)(p|P?)(s|S?))\://)?([a-zA-Z0-9]+[\.]+[a-zA-Z0-9]+[\.]+[a-zA-Z0-9])$
What about this one: (((f|ht)tp://)[-a-zA-Z0-9#:%_\+.~#?&//=]+) ?
This Regular Expression has been tested to work for the following
http|https://host[:port]/[?][parameter=value]*
public static final String URL_PATTERN = "(https?|ftp)://(www\\.)?(((([a-zA-Z0-9.-]+\\.){1,}[a-zA-Z]{2,4}|localhost))|((\\d{1,3}\\.){3}(\\d{1,3})))(:(\\d+))?(/([a-zA-Z0-9-._~!$&'()*+,;=:#/]|%[0-9A-F]{2})*)?(\\?([a-zA-Z0-9-._~!$&'()*+,;=:/?#]|%[0-9A-F]{2})*)?(#([a-zA-Z0-9._-]|%[0-9A-F]{2})*)?";
PS. It also validates on localhost link.
(Thoroughly written by me :-))