Regular Expression, pattern matching in xsd

Question

I was wondering how to make a regular expression for any character except * and + . I\'ve tried ([^*+]) and (\\[^*+]) but

Answer 1

You were close:

[^*+]

Notice that there's no need for escaping those characters inside square brackets, because they have no special meaning there.

EDIT: according to http://www.regular-expressions.info/charclass.html:

Note that the only special characters or metacharacters inside a character class are the closing bracket (]), the backslash (), the caret (^) and the hyphen (-). The usual metacharacters are normal characters inside a character class, and do not need to be escaped by a backslash. To search for a star or plus, use [+*]. Your regex will work fine if you escape the regular metacharacters inside a character class, but doing so significantly reduces readability.

So the problem is not in the character escaping. Maybe you need to match all those occurences. In that case, look up @Cfreak's answer.

Answer 2

In the XML Schema regex flavor, you must not add regex delimiters (i.e., the / at either end of /^[^*+]+$/). You also don't need to use anchors (i.e., the ^ at the beginning and $ at the end); all regex matches are automatically anchored at both ends. That line should read:

<xsd:pattern value="[^*+]+"></xsd:pattern>

...meaning the whole element must consist of one or more of any characters except * and +.

Answer 3

You need to test the entire string. But you don't actually have to escape it in the character class:

/^[^*+]+$/