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I currently have this Regex:
.*[a-z]$|.*[A-Z]$|.*[0-9]$
I need to add the following to the existing - /:.#[]()
Does anyone know how to do it?
You don't need multiple OR conditions. Have only one and include special characters also:
[0-9A-Za-z\/#:.()\[\]]
Per your comments, this should do it for you:
^[a-zA-Z0-9/:.#\[\]()]+$
This will match a string containing only the characters you've specified.
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/edit?o=U&video_id=xxxxxxxxxxxx&show_mt=1
/edit?o=U&show_mt=1&video_id=xxxxxxxxxxxx
I want get video_id query using regex (xxxxxxxxx)
I am using VB.NET
Thanks for help!
M. Deger
Edit: I want only regex code eg: [^?]+(?:\?video_id=([^&]+).*)?
Untested:
(?<=video_id\s*=\s*)[^&]*(?=&|$)
or, result in capture group 1
video_id\s*=\s*([^&]*)(?:&|$)
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what is the regular expression to validate a string. the string should contain
all lower case
_ is allowed in the string.
Thanks in advance.
You need to use regex pattern
^[a-z_]*$
You can use a POSIX class as an alternative to a simple character class to achieve this:
[[:lower:]_]*
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Regex:password all character are allowed,but not include space,length is 8-16.please give me an efftive help.
Consider the following Regex...
^\S{8,16}$
Good Luck!
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I have the following line of XML:
<indexentry><secondaryie>definition, 3/2/4</secondaryie></indexentry>
And I need a regex that matches the above and converts it as below:
ABC3(the first number)/P-2(second number)-4(third number)
How can I do this?
Use this regex:
([0-9]+)/([0-9]+)/([0-9]+)
And from captured groups #1, #2, #3 make your string:
ABC3\1/P-2\2-4\3
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I have a string: "C:/This/is/a/windows/path/file20110312name20120221.csv".
I want to extract all dates (e.g. [0-9]{8}).
Generally, how can I extract sub-strings matching a pattern from a bigger string?
It is advised to supplement your question with a reproducible example (How to make a great R reproducible example?)
Is this what you look for ?
s <- "C:/This/is/a/windows/path/file20110312name20120221.csv"
regmatches(s, gregexpr("[0-9]{8}", s))