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\': \'([^\']*)\'");
$display("This is the output of Regex: %s", wordsRe.getMatchString(1))
See the User Guide details:
getMatchString(m) is always exactly equivalent to calling the range method on the Str object containing the string that was searched:
range(getMatchStart(m), getMatchLength(m))
Related
I'm trying to run a shell script in linux and want to turn this:
/path/to/(\w+)/b/c
into
/path/to/(\w+)/b/(\w+)\.txt
(where \w+ should remain the same as given in input).
I keep getting 'No match found'.
You need to use the capturing group and then use that in your substitution.
\/r\/path\/to\/(\w+).*
Test string
/r/path/to/teststring/b/c
Substitution
/path/to/\1/b/\1\.txt
Result
/path/to/teststring/b/teststring.txt
I have created a regex101 playground for you here
https://regex101.com/r/R0O3OK/1
I have file names in a URL and want to strip out the preceding URL and filepath as well as the version that appears after the ?
Sample URL
Trying to use RegEx to pull, CaptialForecasting_Datasheet.pdf
The REGEXP_EXTRACT in Google Data Studio seems unique. Tried the suggestion but kept getting "could not parse" error. I was able to strip out the first part of the url with the following. Event Label is where I store URL of downloaded PDF.
The URL:
https://www.dudesolutions.com/Portals/0/Documents/HC_Brochure_Digital.pdf?ver=2018-03-18-110927-033
REGEXP_EXTRACT( Event Label , 'Documents/([^&]+)' )
The result:
HC_Brochure_Digital.pdf?ver=2018-03-18-110927-033
Now trying to determine how do I pull out everything after the? where the version data is, so as to extract just the Filename.pdf.
You could try:
[^\/]+(?=\?[^\/]*$)
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
I'm using the following regex to find URLs in a text file:
/http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+/
It outputs the following:
http://rda.ucar.edu/datasets/ds117.0/.
http://rda.ucar.edu/datasets/ds111.1/.
http://www.discover-earth.org/index.html).
http://community.eosdis.nasa.gov/measures/).
Ideally they would print out this:
http://rda.ucar.edu/datasets/ds117.0/
http://rda.ucar.edu/datasets/ds111.1/
http://www.discover-earth.org/index.html
http://community.eosdis.nasa.gov/measures/
Any ideas on how I should tweak my regex?
Thank you in advance!
UPDATE - Example of the text would be:
this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/).
This will trim your output containing trail characters, ) .
import re
regx= re.compile(r'(?m)[\.\)]+$')
print(regx.sub('', your_output))
And this regex seems workable to extract URL from your original sample text.
https?:[\S]*\/(?:\w+(?:\.\w+)?)?
Demo,,, ( edited from https?:[\S]*\/)
Python script may be something like this
ss=""" this is a test http://rda.ucar.edu/datasets/ds117.0/. and I want this to be copied over http://rda.ucar.edu/datasets/ds111.1/. http://www.discover-earth.org/index.html). http://community.eosdis.nasa.gov/measures/). """
regx= re.compile(r'https?:[\S]*\/(?:\w+(?:\.\w+)?)?')
for m in regx.findall(ss):
print(m)
So for the urls you have here:
https://regex101.com/r/uSlkcQ/4
Pattern explanation:
Protocols (e.g. https://)
^[A-Za-z]{3,9}:(?://)
Look for recurring .[-;:&=+\$,\w]+-class (www.sub.domain.com)
(?:[\-;:&=\+\$,\w]+\.?)+`
Look for recurring /[\-;:&=\+\$,\w\.]+ (/some.path/to/somewhere)
(?:\/[\-;:&=\+\$,\w\.]+)+
Now, for your special case: ensure that the last character is not a dot or a parenthesis, using negative lookahead
(?!\.|\)).
The full pattern is then
^[A-Za-z]{3,9}:(?://)(?:[\-;:&=\+\$,\w]+\.?)+(?:\/[\-;:&=\+\$,\w\.]+)+(?!\.|\)).
There are a few things to improve or change in your existing regex to allow this to work:
http[s]? can be changed to https?. They're identical. No use putting s in its own character class
[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),] You can shorten this entire thing and combine character classes instead of using | between them. This not only improves performance, but also allows you to combine certain ranges into existing character class tokens. Simplifying this, we get [a-zA-Z0-9$-_#.&+!*\(\),]
We can go one step further: a-zA-Z0-9_ is the same as \w. So we can replace those in the character class to get [\w$-#.&+!*\(\),]
In the original regex we have $-_. This creates a range so it actually inclues everything between $ and _ on the ASCII table. This will cause unwanted characters to be matched: $%&'()*+,-./0123456789:;<=>?#ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_. There are a few options to fix this:
[-\w$#.&+!*\(\),] Place - at the start of the character class
[\w$#.&+!*\(\),-] Place - at the end of the character class
[\w$\-#.&+!*\(\),] Escape - such that you have \- instead
You don't need to escape ( and ) in the character class: [\w$#.&+!*(),-]
[0-9a-fA-F][0-9a-fA-F] You don't need to specify [0-9a-fA-F] twice. Just use a quantifier like so: [0-9a-fA-F]{2}
(?:%[0-9a-fA-F][0-9a-fA-F]) The non-capture group isn't actually needed here, so we can drop it (it adds another step that the regex engine needs to perform, which is unnecessary)
So the result of just simplifying your existing regex is the following:
https?://(?:[$\w#.&+!*(),-]|%[0-9a-fA-F]{2})+
Now you'll notice it doesn't match / so we need to add that to the character class. Your regex was matching this originally because it has an improper range $-_.
https?://(?:[$\w#.&+!*(),/-]|%[0-9a-fA-F]{2})+
Unfortunately, even with this change, it'll still match ). at the end. That's because your regex isn't told to stop matching after /. Even implementing this will now cause it to not match file names like index.html. So a better solution is needed. If you give me a couple of days, I'm working on a fully functional RFC-compliant regex that matches URLs. I figured, in the meantime, I would at least explain why your regex isn't working as you'd expect it to.
Thanks all for the responses. A coworker ended up helping me with it. Here is the solution:
des_links = re.findall('http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+', des)
for i in des_links:
tmps = "/".join(i.split('/')[0:-1])
print(tmps)
This is probably straightforward but I'm not even sure which phrase I should google to find the answer. Forgive my noobiness.
I've got strings (filenames) that look like this:
site12345678_date20160912_23001_to_23100_of_25871.txt
What this naming convention means is "Records 23001 through 23100 out of 25871 for site 12345678 for September 12th 2016 (20160912)"
What I want to do is extract the date part (those digits between _date and the following _)
The Regex: .*(_date[0-9]{8}).* will return the string _date20160912. But what I'm actually looking for is just 20160912. Obviously, [0-8]{8} doesn't give me what I want in this case because that could be confused with the site, or potentially record counts
How can I responsibly accomplish this sort of 'substringing' with a single regular expression?
You just need to shift you parentheses so as to change the capture group from including '_date' in it. Then you would want to look for your capture group #1:
If done in python, for example, it would look something like:
import re
regex = '.*_date([0-9]{8}).*'
str = 'site12345678_date20160912_23001_to_23100_of_25871.txt'
m = re.match(regex, str)
print m.group(0) # the whole string
print m.group(1) # the string you are looking for '20160912'
See it in action here: https://eval.in/641446
The Regex: .*(_date[0-9]{8}).* will return the string _date20160912.
That means you are using the regex in a method that requires a full string match, and you can access Group 1 value. The only thing you need to change in the regex is the capturing group placement:
.*_date([0-9]{8}).*
^^^^^^^^^^
See the regex demo.
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 :-))