Menu
I don't write many regular expressions so I'm going to need some help on the one.
I need a regular expression that can validate that a string is an alphanumeric comma delimited string.
Examples:
123, 4A67, GGG, 767 would be valid. 12333, 78787&*, GH778 would be invalid fghkjhfdg8797< would be invalid This is what I have so far, but isn't quite right: ^(?=.*[a-zA-Z0-9][,]).*$
Any suggestions?
325
The 0-9 indicates characters 0 through 9, the comma , indicates comma, and the semicolon indicates a ; . The closing ] indicates the end of the character set. The plus + indicates that one or more of the "previous item" must be present.
[] denotes a character class. () denotes a capturing group. [a-z0-9] -- One character that is in the range of a-z OR 0-9.
To split a string by a regular expression, pass a regex as a parameter to the split() method, e.g. str. split(/[,. \s]/) . The split method takes a string or regular expression and splits the string based on the provided separator, into an array of substrings.
\s stands for “whitespace character”. Again, which characters this actually includes, depends on the regex flavor. In all flavors discussed in this tutorial, it includes [ \t\r\n\f]. That is: \s matches a space, a tab, a carriage return, a line feed, or a form feed.
Sounds like you need an expression like this:
^[0-9a-zA-Z]+(,[0-9a-zA-Z]+)*$
Posix allows for the more self-descriptive version:
^[[:alnum:]]+(,[[:alnum:]]+)*$
^[[:alnum:]]+([[:space:]]*,[[:space:]]*[[:alnum:]]+)*$ // allow whitespace
If you're willing to admit underscores, too, search for entire words (\w+):
^\w+(,\w+)*$
^\w+(\s*,\s*\w+)*$ // allow whitespaces around the comma
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
