CODEWITHSUNDEEP
regex
Hammelkeule
Hammelkeule

Reputation: 307

Regular Expression If-Then-Else Condition

I have a field, where a user can search for something. They should be able to search for ^[0-9a-zA-Z*]$.

But it is not possible to search just for the character * (wildcard). There must be at least one other letter or number. a* or *a or 1* or *1 is valid.

So there must be at least one number/letter unequal * for a valid search.

I think it should be realizable with the if/then/else condition. Here is my attempt!

"^(?(?=[*])([0-9a-zA-Z:\]{1,})|([0-9a-zA-Z:\*]*))$"
if character = *
then [0-9a-zA-Z:\]{1,} = user has to enter at least one (or more) characters of the group
else = [0-9a-zA-Z:\*]* = user has to enter 0 or more characters of the group

But it doesn't work...

Upvotes: 1

Views: 224

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626690

You may use

^[0-9a-zA-Z*]*[0-9a-zA-Z][0-9a-zA-Z*]*$

See the regex demo

This regex will match zero or more letters/digits/asterisks, then an obligatory letter or digit, and then zero or more letters/digits/asterisks.

Alternatively, you can require a string to have at least 1 letter or digit:

^(?=.*[0-9a-zA-Z])[0-9a-zA-Z*]+$

See another demo. The ^(?=.*[0-9a-zA-Z]) positive lookahead will require a letter or a digit after zero or more any characters but a newline. It can also be written as ^(?=.*\p{Alnum})[\p{Alnum}*]+$ (with double escaping backslash in Java).

Java demo:

String rx = "(?=.*\\p{Alnum})[\\p{Alnum}*]+";
System.out.println("****".matches(rx));       // FALSE
System.out.println("*a**".matches(rx));       // TRUE

Upvotes: 1

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