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
Related
I'm trying to create a custom filter in Google Analytic to remove the query parts of the url which I don't want to see. The url has the following structure
[domain]/?p=899:2000:15018702722302::NO:::
I would like to create a regex which skips the first 12 characters (that is until:/?p=899:2000), and what ever is going to be after that replace it with nothing.
So I made this one: https://regex101.com/r/Xgbfqz/1 (which could be simplified to .{0,12}) , but I actually would like to skip those and only let the regex match whatever is going to be after that, so that I'll be able to tell in Google Analytics to replace it with "".
The part in the url that is always the same is
?p=[3numbers]:[0-4numbers]
Thank you
Your regular expression:
\/\?p=\d{3}\:\d{0,4}(.*)
Tested in Golang RegEx 2 and RegEx101
It search for /p=###:[optional:####] and capture the rest of the right side string.
(extra) JavaScript:
paragraf='[domain]/?p=899:2000:15018702722302::NO:::'
var regex= /\/\?p=\d{3}\:\d{0,4}(.*)/;
var match = regex.exec(paragraf);
alert('The rest of the right side of the string: ' + match[1]);
Easily use "[domain]/?p=899:2000:15018702722302::NO:::".substr(12)
You can try this:
/\?p\=\d{3}:\d{0,4}
Which matches just this: ?p=[3numbers]:[0-4numbers]
Not sure about replacing though.
https://regex101.com/r/Xgbfqz/1
I am a new user of svlib package in systemverilog environment. Refer to Verilab svlib. I have following sample text , {'PARAMATER': 'lollg_1', 'SPEC_ID': '1G3HSB_1'} and I want to use regex to extract 1G3HSB from this text.
For this reason, I am using the following code snippet but I am getting the whole line instead of only the information.
wordsRe = regex_match(words[i], "\'SPEC_ID\': \'(.*?)\'");
$display("This is the output of Regex: %s", wordsRe.getStrContents())
Can anybody direct me what is going wrong?
The output I am getting : {'PARAMATER': 'lollg_1', 'SPEC_ID': '1G3HSB_1'}
And, I want to get: 1G3HSB_1
It seems you need to get the contents of the first capturing group with getMatchString(1). Also, you need to use a greedy quantifier (lazy ones are not POSIX compliant) and a negated bracket expression - [^']* instead of .*?:
wordsRe = regex_match(words[i], "\'SPEC_ID\': \'([^\']*)\'");
$display("This is the output of Regex: %s", wordsRe.getMatchString(1))
See the User Guide details:
getMatchString(m) is always exactly equivalent to calling the range method on the Str object containing the string that was searched:
range(getMatchStart(m), getMatchLength(m))
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
I moved to a new website and it mangled up my URL's. Now blog posts are accessible from multiple URL's and would like to redirect one pattern to the other.
I am trying to redirect the first case to the second case:
~/blogs/johndoe/john-doe/2014/03/14/test-article1 =>
~/blogs/john-doe/2014/03/14/test-article1
~/blogs/jimjones/jim-jones/2014/03/14/test-articleb =>
~/blogs/jim-jones/2014/03/14/test-articleb
How do I create a pattern smart enough to slice out the first "johndoe" and "jimjones"? I am using this for IIS rewrite but I think any RegEx should work. Thanks for any help.
This works:
^~/blogs/\w+/(\w+)-(\w+)/(\d{4})/(\d\d)/(\d\d)/([\w-]+)$
Debuggex Demo
It just discards the non-dash name. It doesn't know if its equal to the dash name or not. And it also assumes that the date numbers are valid. 9899/45/33 would be matched.
Capture groups:
First name
Last name
Year
Month
Day
Article name
I don't know about IIS rewrites, but this should work:
/^~/blogs\/[a-z]+\/ -> ~/blogs/
The regular expression will match the start of a string, following by ~/blogs/, followed by a string of all lowercase characters.
I don't use IIS, but this should be at least close.
Pattern:
^blogs/\w+/(\w+/)
Action
blogs/{R:1}
Handy usage doc
Can someone assist in creating a Regex for the following situation:
I have about 2000 records for which I need to do a search/repleace where I need to make a replacement for a known item in each record that looks like this:
<li>View Product Information</li>
The FILEPATH and FILE are variable, but the surrounding HTML is always the same. Can someone assist with what kind of Regex I would substitute for the "FILEPATH/FILE" part of the search?
you may match the constant part and use grouping to put it back
(<li>View Product Information</li>)
then you should replace the string with $1your_replacement$2, where $1 is the first matching group and $2 the second (if using python for instance you should call Match.group(1) and Match.group(2))
You would have to escape \ chars if you're using Java instead.