I have a requirement where I need to validate an input field.
It should be alphanumeric characters after 5 characters there should be comma (,)
Example:
K9,d3,dk,33,kd
[a-zA-Z0-9]{5}[,]
but after K9, it gives that Regex pattern not matched
Because your regex says that the "," comma has to be after exactly 5 characters.
But you want a comma after every second char.
Try this:
[a-zA-Z0-9,]{5}
And if it is always 2x chars followed by one comma try this:
([a-zA-Z0-9]{2},)+
This means two chars followed by a comma which can appear one or more times.
And last but not least a shorten form:
(\w{2},)+
Good reference explaining regex in java: http://www.vogella.com/tutorials/JavaRegularExpressions/article.html
Related
Here's the list of my random numbers
r7l5ndecvz9
0qb6rtxsd6ui
dj5iuzpq5vn
rysquf0jkek
435vw5h2qag
And what I want to do is to match or trim the rest after the number of characters (lets say 4), and I want to output as this:
r7l5
0qb6
dj5i
rysq
435v
See, I only want to match the rest of the characters after the 4th character. I tried this regex expression .{4}\b but it only matches the last 4 characters. I also tried this expression ^(\S\S\S\S) but it only matches the first 4 characters. I want to match the rest AFTER the 4th character. Any idea what I'm doing wrong?
Given the sample inputs and outputs, you want to capture the first four characters. The regex for this is ^(.{4})
The caret matches the start of the line, the dot matches any character, the 4 in braces indicates 4 occurrences, and the parentheses capture the result for later access.
Here's working Python code:
import re
inputs = ['r7l5ndecvz9','0qb6rtxsd6ui','dj5iuzpq5vn','rysquf0jkek','435vw5h2qag']
for str in inputs:
match = re.search('^(.{4})',str)
print(match.group(0))
I have trouble understanding why my regex query takes one extra character besides the symbols I have told regex to include into the query, so this is my regex:
([\-:, ]{1,})[^0-9]
This is my test text:
Test- Product-: 1 --- 3 hour ,--kayak:--rental
It always includes the first character of each starting word, like P on Product or h on hour, how can I prevent regex from including those first characters?
I am trying to get all dashes, double points, comma and spaces excluding numbers or any characters.
The [^0-9] part of your regex matches any char but a digit, so you should remove it from your pattern.
There is no need to wrap the character class with a capturing group, and {0,1} is equal to +, so the whole regex can be shortened to
[-:, ]+
Note that - in the initial and end positions inside a character class does not have to be escaped.
I am trying to write a regex to max a sequence of numbers that is 5 digits long or over, but I ignore any spaces, dashes, parens, or hashes when doing that analysis. Here's what I have so far.
(\d|\(|\)|\s|#|-){5,}
The problem with this is that this will match any sequence of 5 characters including those characters I want to ignore, so something like "#123 " would match. While I do want to ignore the # and space character, I still need the number itself to be 5 digits or more in order to qualify at a match.
To be clear, these would match:
1-2-3-4-5
123 45
2(134) 5
Bonus points if the matching begins and ends with a number rather than with one of those "special characters" I am excluding.
Any tips for doing this kind of matching?
If I understood requirements right you can use:
^\d(?:[()\s#-]*\d){4,}$
RegEx Demo
It always matches a digit at start. Then it is followed by 4 or more of a non-capturing group i.e. (?:[()\s#-]*\d) which means 0 or more of any listed special character followed by a digit.
So just repeat a digit, followed by any other sequence of allowed characters 5 or more times:
^(\d[()\s#-]*){5,}$
You can ensure it ends on a digit if you subtract one of the repetitions and add an explicit digit at the end:
^(\d[()\s#-]*){4,}\d$
You can suggest non-digits with \D so et would be something like:
(\d\D*){5,}
Here is a guide.
Using this for an example string
+$43073$7
and need the 5 number sequence from it I'm using the Regex expression
#"\$+(?<lot>\d{5})"
which is matching up any +$ in the string. I tried
#"^\$+(?<lot>\d{5})"
as the +$ are always at the beginning of the string. What will work?
If you use anchor ^, you need to include the + symbol at the first and don't forget to escape it because + is a special meta character in regex which repeats the previous token one or more times.
#"^\+\$(?<lot>\d{5})"
And without the anchor, it would be like
#"\$(?<lot>\d{5})"
And get the 5 digit number you want from group index 1.
DEMO
I would match what you want:
\d+
or if you only want digits after "special" characters at the start of input:
^\W+(\d+)
grabbing group 1
I need a regular expression for 4 characters. The first 3 characters must be a number and the last 1 must be a letter or a digit.
I formed this one, but it not working
^([0-9]{3}+(([a-zA-Z]*)|([0-9]*)))?$
Some valid matches: 889A, 777B, 8883
I need a regular expression for first 3 will be a number and the last 1 will be a alphabet or digit
This regex should work:
^[0-9]{3}[a-zA-Z0-9]$
This assumes string is only 4 characters in length. If that is not the case remove end of line anchor $ and use:
^[0-9]{3}[a-zA-Z0-9]
Try this
This will match it anywhere.
\d{3}[a-zA-Z0-9]
This will match only beginning of a string
^\d{3}[a-zA-Z0-9]
You can also try this website: http://gskinner.com/RegExr/
It makes it very easy to create and test your regex.
Just take the stars out...
^([0-9]{3}+(([a-zA-Z])|([0-9])))?$
The stars mean zero or more of something before it. You are already using an or (|) so you want to match exactly one of the class, or one of the other, not zero or more of the class, or zero or more of the other.
Of course, it can be simplified further:
^\d{3}[a-zA-Z\d]$
Which literally means... three digits, followed by a character from either lowercase or uppercase a-z or any digit.