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I have some URLs:
google.com
stack.bing.com
yahoo.com/text/4378
yahoo.com/65456/4378/76576
How to remove URL with more than 2 / characters? After removing, it only has:
google.com
stack.bing.com
How to do it with regular expressions?
In this link http://textmechanic.com/Remove-Lines-Containing.html, it has Enable regular expression search function. So, i want to use regular expression for it.
You can use this regex for matching URLs with less than 2 slashes excluding cases of http://example):
^(?!.*?\/[^\/\n]+\/).+$
RegEx Demo
Or you can inverse the regex for removal:
^(?=.*?\/[^\/\n]+\/).+$
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I have the following data in my notepad++ file:
aa(40)
aa(50)
aa(4)
aa(8)
aa(40)
How do I search for it using regular expressions in notepad++?
Note : aa is always constant
In Notepad++:
Go to Search >> Find
Select Regular Expressions
Enter following regex: ^aa\(\d+\)
Find All in All Opened Documents
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does anyone know how to write the regex for this list of urls for workbox
/
/tracker/patientlist
/tracker/visitlist
/tracker/triage
i am new to regex
Assuming you want to match the root url /, or /tracker/*, try this:
/(tracker/\w+)?
Here is one way.(Not sure what other URI variations you may have, but this should do it). If the first part is always /tracker, then use Bohemian's answer.
(\/[a-zA-Z]+)
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I have a pattern like [word1][wor2]. I want to replace to {#link word2|word1}, word1, word2 could be anything.
I am very bad on using regular expression. Could anyone help me use regular expression match the pattern and replace to new pattern.
Thanks in advance.
Try the following find and replace, in regex mode:
Find: \[([^\]]+)\]\[([^\]]+)\]
Replace: {#link $2|$1}
Demo
The uses the approach of capturing the two adjacent word in square brackets, and then replacing with the text you want.
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I could anybody help me out?
I am looking for some regexp to use to validate US Zip codes and Canadian Postal codes in AngularJS
This should work for you:
^(\d{5}(-\d{4})?|[A-Z]\d[A-Z] *\d[A-Z]\d)$
http://regex101.com/r/nO4zF5
Remove the ^ and $ anchors if you don't want to match the start and end of the string.
When using in Javascript remember that you need to use the / delimiter:
if (/^(\d{5}(-\d{4})?|[A-Z]\d[A-Z] *\d[A-Z]\d)$/.test(str)) {
// it's a match!
}
And directly, as an Angular ng-pattern:
ng-pattern="/^(\d{5}(-\d{4})?|[A-Z]\d[A-Z] *\d[A-Z]\d)$/"
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I would like to search for everything before the string http://thepaperwall.com/wallpapers/ but not including that string.
I would also like to search for everything after .jpg but not including .jpg.
So my final string should look like this:
http://thepaperwall.com/wallpapers/cityscape/small/small_642331afcd78d0840485bb352d99b289b50e8467.jpg
How can I do this?
You can use look-behind and look-ahead to get a string position between 2 other strings:
(?<=http://thepaperwall.com/wallpapers/).*(?=\.jpg)
You need a non-capturing group:
(.*)(?:http:\/\/thepaperwall\.com\/wallpapers\/.*\.jpg)(.*)
Live demo