I have a laravel validation rule something like this
public function rules(){
return [
'title' => 'required|max:100',
'featured_image' => 'required|max:100|regex:(\d)+.(?:jpe?g|png|gif)',
];
}
I have a txt field where i dynamically add an image name, something like this (8123123123.jpg OR 234234234.png). If the text field doesn't have this pattern i want to show an error.
Now this regex does work in http://regexr.com/ but in laravel it doesn't. So basically it should look for digits as file name and should end with .jpg or .png
Any help is appreciated
use like this
'featured_image' => ['required', 'max:100', 'regex:/(\d)+.(?:jpe?g|png|gif)/']
You have to add a regex delimiter http://php.net/manual/en/regexp.reference.delimiters.php
According to the docs you should put your validations in a array:
regex:pattern
The field under validation must match the given regular expression.
Note: When using the regex pattern, it may be necessary to specify rules in an array instead of using pipe delimiters, especially if the regular expression contains a pipe character.
Related
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
Okay so I need to create an advance filter in Google Analytics that includes "breast", but DOES NOT include "before" "after" or "blog" in the url. I also want to filter out .jpg file extensions.
Here are example URLs that I want the filter to return:
http://www.doctortaylor.com/breast-lift-surgery/
http://www.doctortaylor.com/breast-augmentation-pasadena-and-los-angeles-area/
I want to filter out any urls that are before and after photo pages, and any actual .jpg file urls.
I'm a regex beginner, but this is pretty advanced. Any help would be greatly appreciated!!
This regular expression does fairly well:
^(?!before|after|blog)*((?!before|after|blog).)*breast(?!before|after|blog|\.jpg)*((?!before|after|blog|\.jpg).)*$
UPDATED: I have updated the expression to capture all scenarios, even characters that begin or end the string. This regular expression excludes all words that you list in your description and correctly identifies the word breast.
MATCHES
http://www.doctortaylor.com/breast-lift-surgery/
http://www.doctortaylor.com/breast-augmentation-pasadena-and-los-angeles-area/
DOES NOT MATCH
http://www.doctortaylor.com/breast-lift-surgeryblog/
http://www.doctortaylor.com/breast-lift-surgery.jpg/
http://blog.doctortaylor.com/breast-lift-surgery/
http://www.doctortaylor.com/after-breast-lift-surgery/
This regular expression uses an equivalent of inverse matching.
When I post a value from my page an extra string is create and I would like to remove the 'pattern = "' is there a tag i could use or replace function i can use to remove pattern =" and the ending ". Please find below scenario:
pattern = "apple"
Desired output
apple
I tried using but to no success.Is there another method i could use newbie at django python.
{{ pattern|split }}
Write a custom templatetag and use a regular expression (a capture group would do the trick) to replace the desired part.
If the quotation marks are added to your string when you are submitting the form, you could use the .strip() method when accessing the POST data. You can also specify what characters you want to remove. Check it out here: http://docs.python.org/2/library/stdtypes.html
I have the following strings in my application.
/admin/stylesheets/11
/admin/javascripts/11
/contactus
what I want to do is to write a regular expression to capture anything other than string starting with 'admin'
basically my regex should capture only
/contactus
by excluding both
/admin/stylesheets/11
/admin/javascripts/11
to capture all i wrote
/.+/
and i wrote /(admin).+/ which captures everything starts with 'admin'. how can i do the reverse. I mean get everything not starting with 'admin'
thanks in advance
cheers
sameera
EDIT - Thanks all for the answers
I'm using ruby/ Rails3 and trying to map a route in my routes.rb file
My original routes file is as followss
match '/:all' => 'page#index', :constraints => { :all => /.+/ }
and i want the RegEx to replace /.+/
thanks
If the language/regular expression implementation you are using supports look-ahead assertions, you can do this:
^/(?!admin/).+/
Otherwise, if you only can use basic syntax, you will need to do something like this:
^/([^a].*|a($|[^d].*|d($|[^m].*|m($|[^i].*|i($|[^n].*)))))
I am trying to write a pattern for extracting the path for files found in img tags in HTML.
String string = "<img src=\"file:/C:/Documents and Settings/elundqvist/My Documents/My Pictures/import dialog step 1.JPG\" border=\"0\" />";
My Pattern:
src\\s*=\\s*\"(.+)\"
Problem is that my pattern will also include the 'border="0" part of the img tag.
What pattern would match the URI path for this file without including the 'border="0"?
Your pattern should be (unescaped):
src\s*=\s*"(.+?)"
The important part is the added question mark that matches the group as few times as possible
This one only grabs the src only if it's inside of an tag and not when it is written anywhere else as plain text. It also checks if you've added other attributes before or after the src attribute.
Also, it determines whether you're using single (') or double (") quotes.
\<img.+src\=(?:\"|\')(.+?)(?:\"|\')(?:.+?)\>
So for PHP you would do:
preg_match("/\<img.+src\=(?:\"|\')(.+?)(?:\"|\')(?:.+?)\>/", $string, $matches);
echo "$matches[1]";
for JavaScript you would do:
var match = text.match(/\<img.+src\=(?:\"|\')(.+?)(?:\"|\')(?:.+?)\>/)
alert(match[1]);
Hopefully that helps.
Try this expression:
src\s*=\s*"([^"]+)"
I solved it by using this regex.
/<img.*?src="(.*?)"/g
Validated in https://regex101.com/r/aVBUOo/1
You want to play with the greedy form of group-capture. Something like
src\\s*=\\s*\"(.+)?\"
By default the regex will try and match as much as possible
I am trying to write a pattern for extracting the path for files found in img tags in HTML.
Can we have an autoresponder for "Don't use regex to parse [X]HTML"?
Problem is that my pattern will also include the 'border="0" part of the img tag.
Not to mention any time 'src="' appears in plain text!
If you know in advance the exact format of the HTML you're going to be parsing (eg. because you generated it yourself), you can get away with it. But otherwise, regex is entirely the wrong tool for the job.
I'd like to expand on this topic as usually the src attribute comes unquoted so the regex to take the quoted and unquoted src attribute is:
src\s*=\s*"?(.+?)["|\s]