My vehicle number is something like this.
1-0001,
10-0010,
111-000
A-3049,
KJ-6825,
AAC-3422
The following expression is the one I found.
^([a-zA-Z]{1,3}|([0-9]{1,3}))-[0-9]{4}
But I want that first three characters should not be all zeros or last four characters not be all zeros how do I make a valid expression ?
^([a-zA-Z]{1,3}|((?!0*-)[0-9]{1,3}))-[0-9]{4}(?<!0{4})
Try this.See demo.
https://regex101.com/r/mS3tQ7/12
The lookahead and lookbehind will make sure that there are no all 0\s at start or all 0's at end.
Also use
^(?>[a-zA-Z]{1,3}|(?!0*-)[0-9]{1,3})-[0-9]{4}(?<!0{4})
To make sure you dont make partial matches.
https://regex101.com/r/mS3tQ7/13
How about this:
^((?!0000)([a-zA-Z]{1,3}|([0-9]{1,3}))-[0-9])|(([a-zA-Z]{1,3}|([0-9]{1,3}))-[0-9])(?!000)
Debuggex Demo
And what about this:
^([0-9]{1,3}|[A-Z]{1,3})-[0-9]{1,4}
Related
I have the following string:
"121 fd412 4151 3213, 421, 423 41241 fdsfsd"
And I need to get 3213 and 421 - because they both have space in front of them, and a coma behind.
The result will be set inside the string array...How can I do that?
"\\d+" catches every integer.
"\s\\d+(,)" throws some memory errors.
EDIT.
space to the left (<-) of the number, coma to the right (->)
EDIT 2.
string mainString = "Tests run: 5816, 8346, 28364 iansufbiausbfbabsbo3 4";
MatchCollection c = Regex.Matches(a, #"\d+(?=\,)");
var myList = new List<String>();
foreach(Match match in c)
{
myList.Add(match.Value);
}
Console.Write(myList[1]);
Console.ReadKey();
Your regex syntax is incorrect for wanting to match both digits, if you want them as separate results, you could do:
#"\s(\d+),\s(\d+)\s"
Live Demo
Edit
#"\s(\d+),"
Live Demo
\s\\d+(,):
\s is not properly escaped, should be \\s, same as for \\d
\\d matches single digit, you need \\d+ - one or more consecutive digits
(,) captures comma, do you really need this? seems like you need to capture a number, so \\s(\\d+),
you said "because they both have space behind them, and a coma in front", so probably ,\\s(\\d+)
How about this expression :
" \d+," // expression without the quotes
it should find what you need.
How to work with regular expression can you check on the MSDN
Hope it helps
Another solution
\s(\d+), // or maybe you'll need a double slash \\
Output:
3213
421
Demo
I think you mean you're looking for something like ,<space><digit> not ,<digit><space>
If so, try this:
, (\d+) //you might need to add another backslash as the others have noted
Well, based on your new edit
\s(\d+),
Test it here
It's all you need, only the numbers
\d+(?=\,)
I need a regular expression validation expression that will
ALLOW
positive number(0-9)
, and .
DISALLOW
letter(a-z)
any other letter or symbol except . and ,
for example, on my asp.net text box, if I type anything#!#--, the regular expression validation will disallow it, if I type 10.000,50 or 10,000.50 it should allowed.
I've been trying to use this regex:
^\d+(\.\d\d)?$
but my textbox also must allow , symbol and I tried using only integer regex validation, it did disallow if I type string, but it also disallow . and , symbol while it should allow number(0-9) and also . and , symbol
Don't Use \d to match [0-9] in .NET
First off, in .NET, \d will match any digits in any script, such as:
654۳۲١८৮੪૯୫୬१७੩௮௫౫೮൬൪๘໒໕២៧៦᠖
So you really want to be using [0-9]
Incomplete Spec
You say you want to only allow "digits, commas and periods", but I don't think that's the whole spec. That would be ^[0-9,.]+$, and that would match
...,,,
See demo.
Tweaking the Spec
It's hard to guess what you really want to allow: would 10,1,1,1 be acceptable?
We could start with something like this, to get some fairly well-formed strings:
^(?:[0-9]+(?:[.,][0-9]+)?|[1-9][0-9]{0,2}(?:(?:\.[0-9]{3})*|(?:,[0-9]{3})*)(?:\.[0-9]+)?)$
Play with the demo, see what should and shouldn't match... When you are sure about the final spec, we can tweak the regex.
Sample Matches:
0
12
12.123
12,12
12,123,123
12,123,123.12456
12.125.457.22
Sample Non-Matches:
12,
123.
1,1,1,1
Your regex would be,
(?:\d|[,\.])+
OR
^(?:\d|[,\.])+$
It matches one or more numbers or , or . one or more times.
DEMO
Maybe you can use this one (starts with digit, ends with digit):
(\d+[\,\.])*\d+
If you need more sophisticated price Regex you should use:
(?:(?:[1-9]\d?\d?([ \,\.]?\d{3})*)|0)(?:[\.\,]\d+)?
Edit: To make it more reliable (and dont get 00.50) you can add starting and ending symbol check:
(^|\s)(?:(?:[1-9]\d?\d?([ \,\.]?\d{3})*)|0)(?:[\.\,]\d+)($|\s)?
I think the best regex for your condition will be :
^[\d]+(?:,\d+)*(?:\.\d+)?$
this will validate whatever you like
and at the same time:
not validate:
numbers ending in ,
numbers ending in .
numbers having . before comma
numbers having more than one decimal points
check out the demo here : http://regex101.com/r/zI0mJ4
Your format is a bit strange as it is not a standard format.
My first thought was to put a float instead of a string and put a Range validation attribute to avoid negative number.
But because of formatting, not sure it would work.
Another way is the regex, of course.
The one you propose means :
"some numbers then possibly a group formed by a dot and two numbers exactly".
This is not what you exepected.
Strictly fitted your example of a number lower than 100,000.99 one regex could be :
^[0-9]{1-2}[\.,][0-9]{3}([\.,][0-9]{1-2})?$
A more global regex, that accept all positive numbers is the one posted by Avinash Raj : (?:\d|[,\.])+
I have a textbox and in it a value like $8.00 I want to validate this textbox to always check for amount values and not accept letters or anything other than a value in the format of 0.00. How can I achieve this in a RegularExpressionValidator?
Thank you for the help.
The RegEx you are looking for is #"^\d+\.\d\d"
It matches strings with 1+ digits before point and exactly two digits after
If you want it to allow start a string from $, then use #"^\$\d+\.\d\d" or #"^\$?\d+\.\d\d" for optional $.
If you want $ to be separated from digits with spaces then use #"^\$?\s*\d+\.\d\d"
The following regular expressiong will allow numbers in the following format (12345.67, 0, 0.1)
^\d{1,5}(.\d{1,2})?$
I used one of the following before i hope it helps try it.
\d{1,3}(.)\d{1,2}
or
\d{1,3}.\d{2}
I have the two following strings
/signon/default.wl?rs=WLW11.10&vr=2.0&fn=_top
/signon/default.wl?fn=%5Ftop&newdoor=true&rs=WLW11%2E10&vr=2%2
I would like to match all the strings except the ones that do not contain newdoor
so far i have the following regex
/signon/default.wl\?(?=[\w]*)(?!newdoor)
but it matches all strings.
can anyone point out what im doing wrong.
You can try this /signon/default.wl\?(?!.*newdoor).*
It asserts using negative lookahead that there is no occurrence of newdoor in the input string. Code will look like this
resultString = Regex.Match(subjectString, #"/signon/default.wl\?(?!.*newdoor).*",
RegexOptions.IgnoreCase).Value;
/signon/default.wl\?(?=[\w]*)(?!newdoor)
Both (?=) and (?!) are zero-width assertions that don't consume any input. Try:
/signon/default.wl\?(.(?!newdoor))*$
why not use something like:
subjectString.contains("newdoor");
I am looking for a way to get words out of a sentence. I am pretty far with the following expression:
\b([a-zA-Z]+?)\b
but there are some occurrences that it counts a word when I want it not to. E.g a word followed by more than one period like "text..". So, in my regex I want to have the period to be at the end of a word zero or one time. Inserting \.? did not do the trick, and variations on this have not yielded anything fruitful either.
Hope someone can help!
A single dot means any character. You must escape it as
\.?
Maybe you want an expression like this:
\w+\.?
or
\p{L}+\.?
You need to add \.? (and not .?) because the period has special meaning in regexes.
to avoid a match on your example "test.." you ask for you not only need to put the \.? for checking first character after the word to be a dot but also look one character further to check the second character after the word.
I did end up with something like this
\w{2,}\.?[^.]
You should also consider that a sentence not always ends with a . but also ! or ? and alike.
I usually use rubulator.com to quick test a regexp