I'm getting a value like this:
myimage.jpg123456jpg
and I need to remove everything after .jpg
how can I write this in razor?
I don't know anything about razor but this regex would match the part you'd like to save in the first result group:
(.+\.jpg)
You can see it in action here: http://regexr.com?2v7ki
Just match on .+\.jpg, which will give you the myimage.jpg section of the text.
Related
I'm trying to solve a regex riddle. Let's say I have rows of hrefs looking like this:
anchor1.in
an3.php
setup.exe
What I want the regex (or any other solution) to do is to take the href title and copy it over to the actual url with a foward slash in front of it.
A successful result would become:
anchor1.in
an3.php
setup.exe
If you can solve this please explain how you did it.
You can use the following to match:
(<a\s+href=")(.*?)(">)(.*?)(<\/a>)
And replace with:
\1\2/\4\3\4\5
See DEMO and Explanation
I am trying to write a regex which will strip away the rest of the path after a particular folder name.
If Input is:
/Repository/Framework/PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces/IDemoReader.cs
Output should be:
/Repository/Framework/PITA/branches/ChangePack-6a7B6
Some constrains:
ChangePack- will be followed change pack id which is a mix of numbers or alphabets a-z or A-Z only in any order. And there is no limit on length of change pack id.
ChangePack- is a constant. It will always be there.
And the text before the ChangePack can also change. Like it can also be:
/Repository/Demo1/Demo2/4.3//PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces
My regex-fu is bad. What I have come up with till now is:
^(.*?)\-6a7B6
I need to make this generic.
Any help will be much appreciated.
Below regex can do the trick.
^(.*?ChangePack-[\w]+)
Input:
/Repository/Framework/PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces/IDemoReader.cs
/Repository/Demo1/Demo2/4.3//PITA/branches/ChangePack-6a7B6/core/src/Pita.x86.Interfaces
Output:
/Repository/Framework/PITA/branches/ChangePack-6a7B6
/Repository/Demo1/Demo2/4.3//PITA/branches/ChangePack-6a7B6
Check out the live regex demo here.
^(.*?ChangePack-[a-zA-Z0-9]+)
Try this.Instead of replace grab the match $1 or \1.See demo.
https://regex101.com/r/iY3eK8/17
Will you always have '/Repository/Framework/PITA/branches/' at the beginning? If so, this will do the trick:
/Repository/Framework/PITA/branches/\w+-\w*
Instead of regex you could can use split and join functions. Example python:
path = "/a/b/c/d/e"
folders = path.split("/")
newpath = "/".join(folders[:3]) #trims off everything from the third folder over
print(newpath) #prints "/a/b"
If you really want regex, try something like ^.*\/folder\/ where folder is the name of the directory you want to match.
I would like to get super scripted text via following html string.
testing to <sup>supers</sup>cript o<sup>n</sup>e
The result I would like to get is like below
supers
n
This is what I tried right now
But the result is not what I want.
<sup>supers
<sup>n
Could anyone give me suggestion please?
You can use lookbehind in your regex:
(?<=<sup>)[^<]*
Update Demo
Use this if there may be other HTML tags between <sup> and </sup>:
(?<=<sup>)(.*?)(?=<\/sup>)
Check the demo.
You were close, just not capturing your match:
Updated regex
(?:<sup>)([^<]*) I just added a capture group around your match
(?<=<sup>)([^<]*?)(?=<\/)
This should work.
See demo.
http://regex101.com/r/sA7pZ0/13
I'm edit an rst file for a document. There are lot of image links I have to edit them one by one, I'd like to ask, is there anybody can help write a regular expression that can transfer it in one time.
The original text looks like:
*Figure 1.2: Where is the dog?* <dog.html#fig_dog>
I'd like it translate it to:
:ref:fig_dog
And there is another one:
*How are you* <how_are_you.html>
I'd like it translate it to:
:ref:how_are_you
I have try some expression in editplus or notepad++, but i can't match them very well.
Search:
\*.*?\*\s*<(?:.*#)?([^.>]+)(\.[^>]*)?>
Replace:
:ref:\1
split into two regexes
To match before the html
<(.*).html.*?>
To match the anchor
<.*.html#(.*?)>
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]