I am trying to create a regEx that will have an optional '+' as the first character and then after accept all digits, making sure no spaces are allowed.
I have tried the following, but it does not seem to work :
new RegExp(/^\+?\d $/);
Seems like you want something like this,
new RegExp(/^\+?\d+$/);
Replace space in your regex with +. \d+ matches one or more digits.
Related
The pattern I need is below:
MA should be the first two characters of a string
The third character should be hyphen (-)
Characters 4 to 10 can be any numeric numbers (0-9)
The eleventh character should be hyphen (-)
Characters 12 to 15 can be any numeric numbers (0-9)
Example:
MA-1234567-1234
I have tried this:
/^(MA*)[0-9]{7}([0-9]{4})$/
The Regex you were using is missing the dashes between character sets, try this:
/^MA-\d{7}-\d{4}$/
Note that to incorporate this with an input box you would need to test against this Regex on keyup, something like this:
var re = /^MA-\d{7}-\d{4}$/;
$('#code').keyup(function() {
if (re.test(this.value)) {
console.log('The code is valid...');
}
});
Example fiddle
Nearly there you were missing dashes for separators and you don't need the paranthesis if you don't want to extract any of the information.
Correct regex is:
/^MA-[0-9]{7}-[0-9]{4}$/
Edit to satisfy new requirements
To match both: CW291291 and MA-1234567-1234
Use the pipe symbol '|' with both patterns to match either one:
/^CW[\d]{6}|MA-[\d]{7}-[\d]{4}$/
I have a regex that accepts payments and amounts below. Allows for formats with "$" "," and two decimal spaces. I need this regex to allow trailing spaces after the decimal places like so '$50.00 '
/^\$?[0-9 ][0-9\, ]*(\.\d{1,2})?$|^\$?[\.]([\d][\d])$/
Thoughts?
\s* will capture any whitespace, including tabs and newlines:
/^\$?[0-9 ][0-9\, ]*(\.\d{1,2})?$|^\$?[\.]([\d][\d])\s*$/
If you're just trying to validate dollar values, I would make it a little simpler (although I don't know your exact use case):
/^\$?[0-9,]+(\.\d{2})?\s*$/
Just adding a [ ]* to the end should do the trick:
/^\$?[0-9 ][0-9\, ]*(\.\d{1,2})?[ ]*$|^\$?[\.]([\d][\d])[ ]*$/
But you can simplify this by wrapping your alternatives in a group. Also note that you don't need to escape , or . inside a character class, and you don't need a character class at all if there is only character you want to match.
/^\$?(?:[0-9 ][0-9, ]*(\.\d{1,2})?|\.([\d][\d]))[ ]*$/
Hello I am trying to create a regex that recognizes money and numbers being inputted. I have to allow numbers because I am expecting non-formatted numbers to be inputted programmatically and then I will format them myself. For some reason my regex is allowing a one letter character as a possible input.
[\$]?[0-9,]*\.[0-9][0-9]
I understand that my regex accepts the case where multiple commas are added and also needs two digit after the decimal point. I have had an idea of how to fix that already. I have narrowed it down to possibly the *\. as the problem
EDIT
I found the regex expression that worked [\$]?([0-9,])*[\.][0-9]{2} but I still don't know how or why it was failing in the first place
I am using the .formatCurrency() to format the input into a money format. It can be found here but it still allows me to use alpha characters so i have to further masked it using the $(this).inputmask('Regex', { regex: "[\$]?([0-9,])*[\.][0-9]{2}" }); where input mask is found here and $(this) is a reference to a input element of type text. My code would look something like this
<input type="text" id="123" data-Money="true">
//in the script
.find("input").each(function () {
if ($(this).attr("data-Money") == "true") {
$(this).inputmask('Regex', { regex: "[\$]?([0-9,])*[\.][0-9]{2}" });
$(this).on("blur", function () {
$(this).formatCurrency();
});
I hope this helps. I try creating a JSfiddle but Idk how to add external libraries/plugin/extension
The "regular expression" you're using in your example script isn't a RegExp:
$(this).inputmask('Regex', { regex: "[\$]?([0-9,])*[\.][0-9]{2}" });
Rather, it's a String which contains a pattern which at some point is being converted into a true RegExp by your library using something along the lines of
var RE=!(value instanceof RegExp) ? new RegExp(value) : value;
Within Strings a backslash \ is used to represent special characters, like \n to represent a new-line. Adding a backslash to the beginning of a period, i.e. \., does nothing as there is no need to "escape" the period.
Thus, the RegExp being created from your String isn't seeing the backslash at all.
Instead of providing a String as your regular expression, use JavaScript's literal regular expression delimiters.
So rather than:
$(this).inputmask('Regex', { regex: "[\$]?([0-9,])*[\.][0-9]{2}" });
use
$(this).inputmask('Regex', { regex: /[\$]?([0-9,])*[\.][0-9]{2}/ });
And I believe your "regular expression" will perform as you expect.
(Note the use of forward slashes / to delimit your pattern, which JavaScript will use to provide a true RegExp.)
Firstly, you can replace '[0-9]' with '\d'. So we can rewrite your first regex a little more cleanly as
\$?[\d,]*\.\d\d
Breaking this down:
\$? - A literal dollar sign, zero or one
[\d,]* - Either a digit or a comma, zero or more
\. - A literal dot, required
\d - A digit, required
\d - A digit, required
From this, we can see that the minimum legal string is \.\d\d, three characters long. The regex you gave will never validate against any one character string.
Looking at your second regex,
[\$]? - A literal dollar sign, zero or one
([0-9,])* - Either a digit or a comma, subexpression for later use, zero or more
[\.] - A literal dot, required
[0-9]{2} - A digit, twice required
This has the exact same minimum matchable string as above - \.\d\d.
edit: As mentioned, depending on the language you may need to escape forward slashes to ensure they aren't misinterpretted by the language when processing the string.
Also, as an aside, the below regex is probably closer to what you need.
[A-Z]{3} ?(\d{0,3}(?:([,. ])\d{3}(?:\2\d{3})*)?)(?!\2)[,.](\d\d)\b
Explanation:
[A-Z]{3} - Three letters; for an ISO currency code
? - A space, zero or more; for readability
( - Capture block; to catch the integer currency amount
\d{0,3} - A digit, between one and three; for the first digit block
(?: - Non capturing block (NC)
([,. ]) - A comma, dot or space; as a thousands delimiter
\d{3} - A digit, three; the first possible whole thousands
(?: - Non capturing block (NC)
\2 - Match 2; the captured thousands delimiter above
\d{3} - A digits, three
)* - The above group, zero or more, i.e. as many thousands as we want
)? - The above (NC) group, zero or one, ie. all whole thousands
) - The above group, i.e everything before the decimal
[.,] - A comma or dot, as a decimal delimiter
(\d{2}) - Capture, A digit, two; ie. the decimal portion
\b - A word boundry; to ensure that we don't catch another
digit in the wrong place.
The negative lookahead was provided by an answer from John Kugelman in this question.
This correctly matches (matches enclosed in square brackets):
[AUD 1.00]
[USD 1,300,000.00]
[YEN 200 000.00]
I need [USD 1,000,000.00], all in non-sequential bills.
But not:
GBP 1.000
YEN 200,000
Having the following regex: ([a-zA-Z0-9//._-]{3,12}[^//._-]) used like pattern="([a-zA-Z0-9/._-]{3,12}[^/._-])" to validate an HTML text input for username, I wonder if is there anyway of telling it to check that the string has only one of the following: ., -, _
By that I mean, that I'm in need of regex that would accomplish the following (if possible)
alex-how => Valid
alex-how. => Not valid, because finishing in .
alex.how => Valid
alex.how-ha => Not valid, contains already a .
alex-how_da => Not valid, contains already a -
The problem with my current regex, is that for some reason, accepts any character at the end of the string that is not ._-, and can't figure it out why.
The other problem, is that it doesn't check to see that it contains only of the allowed special characters.
Any ideas?
Try this one out:
^(?!(.*[.|_|-].*){2})(?!.*[.|_|-]$)[a-zA-Z0-9//._-]{3,12}$
Regexpal link. The regex above allow at max one of ., _ or -.
What you want is one or more strings containing all upper, lower and digit characters
followed by either one or none of the characters in "-", ".", or "_", followed by at least one character:
^[a-zA-Z0-9]+[-|_|\.]{0,1}[a-zA-Z0-9]+$
Hope this will work for you:-
It says starts with characters followed by (-,.,_) and followed and end with characters
^[\w\d]*[-_\.\w\d]*[\w\d]$
Seems to me you want:
^[A-Za-z0-9]+(?:[\._-][A-Za-z0-9]+)?$
Breaking it down:
^: beginning of line
[A-Za-z0-9]+: one or more alphanumeric characters
(?:[\._-][A-Za-z0-9]+)?: (optional, non-captured) one of your allowed special characters followed by one or more alphanumeric characters
$: end of line
It's unclear from your question if you wanted one of your special characters (., -, and _) to be optional or required (e.g., zero-or-one versus exactly-one). If you actually wanted to require one such special character, you would just get rid of the ? at the very end.
Here's a demonstration of this regular expression on your example inputs:
http://rubular.com/r/SQ4aKTIEF6
As for the length requirement (between 3 and 12 characters): This might be a cop-out, but personally I would argue that it would make more sense to validate this by just checking the length property directly in JavaScript, rather than over-complicating the regular expression.
^(?=[a-zA-Z0-9/._-]{3,12}$)[a-zA-Z0-9]+(?:[/._-][a-zA-Z0-9]+)?$
or, as a JavaScript regex literal:
/^(?=[a-zA-Z0-9\/._-]{3,12})[a-zA-Z0-9]+(?:[\/._-][a-zA-Z0-9]+)?$/
The lookahead, (?=[a-zA-Z0-9/._-]{3,12}$), does the overall-length validation.
Then [a-zA-Z0-9]+ ensures that the name starts with at least one non-separator character.
If there is a separator, (?:[/._-][a-zA-Z0-9]+)? ensures that there's at least one non-separator following it.
Note that / has no special meaning in a regex. You only have to escape it if you're using a regex literal (because / is the regex delimiter), and you escape it by prefixing with a backslash, not another forward-slash. And inside a character class, you don't need to escape the dot (.) to make it match a literal dot.
The dot in regex has a special meaning: "any character here".
If you mean a literal dot, you should escape it to tell the regex parser so.
Escape dot in a regex range
I am not very experienced with regex and I need to validate phone numbers using javascript.
I have a textbox which need to be allowed to accept multiple phone numbers with a delimiter of ';' and the characters that can be allowed for the phone numbers are
Numbers
'+'
'-'
Could someone help me on how I can acheive this using javascript and regex/ regular expressions?
Example:
+91-9743574891;+1-570-456-2233;+66-12324576
I tried the following:
^[0-9-+;]+$
Am not sure if this is correct.
You have placed - in wrong place so, your regex is not working.
Try this(your RegEx, but slightly modified):
^[0-9+;-]+$
or
^[-0-9+;]+$
To include a hyphen within a character class then you must do one of the following:
escape the hyphen and use \-,
place hyphen either at the beginning or at the end of the character class.
As the hyphen is used for specifying a range of characters. So, regex engine understands [0-9-+;]+ match any of the characters between 0 to 9, 9 to +(all characters having decimal code-point 57[char 9] to 43[char +] and it fails) and ;.
To be a bit more restrictive, you could use the following regexp:
/^\+[0-9]+(-[0-9]+)+(;\+[0-9]+(-[0-9]+)+)*$/
What it will match:
+91-9743574891
+1-570-456-2233;+66-12324576
What it won't match:
91-9743574891
+15704562233
6612324576
How about this ^([0-9\-\+]{5,15};?)+$
Explanation:
^ #Match the start of the line
[0-9\-\+] #Allow any digit or a +/- (escaped)
{5,15} #Length restriction of between 5 and 15 (change as needed)
;? #An optional semicolon
+ #Pattern can be repeat once or more
$ #Until the end of the line
Only as restrictive as specified could be tighter, See it working here.
Your regex will match what you allow, but I would be a bit more restrictive:
^\+?[0-9-]+(?:;\+?[0-9-]+)*$
See it here on Regexr
That means match an optional "+" followed by a series of digits and dashes. Then there can be any amount of additional numbers starting with a semicolon, then the same pattern than for the first number.