Java Help Regular Expression

[geforce]

"Only letters allowed and at most one blank, which cannot be the first or last character"
What regular expression would meet this and why?

[Fubarable]

How about giving it a try and showing us your attempt? If you need help, please check these links on regular expressions:

Lesson: Regular Expressions: for the basic concepts
Regex Tutorial: for more advanced concepts

[geforce]

I was thinking:

regex = "\w{1,}\s{0,1}\w{1,}" but I keep getting illegal escape character

[geforce]

Since it's only letters it's ^[A-Z-a-z]

[Fubarable]

In Java, your backslashes need to be escaped -- "\\w" not "\w". But even so, your regex won't work. I honestly don't know the solution, but am trying to work on it myself...

[geforce]

Yeah, this one isn't easy.

[Fubarable]

OK, I figured it out. Not that hard actually.

[Fubarable]

What do you think should be the first term of the regex?

[geforce]

Something near "^[\p{Z,1}][\w*]^[\p{Z,1]"

[Fubarable]

No, you're doing what I was doing -- making things too complicated; it's much simpler than this.

The first bit of the regex, the start should be "\\w", that's it. Then after this consider using a vertical bar or pipe to allow the regex to accept one of two following Strings -- either a String made of only characters, or a String with a variable number of characters, a space, and followed by a variable number of characters.

So the regex will look something like "\\w(...|...)" where you would have to fill in the ... parts.

[geforce]

Something near "^[\\w*]\\s{0,1}[\\w*]"

[Fubarable]

That idea can work too, but don't forget that first \\w that must be present. Your code would allow a String that starts with a space, and still allows any white-space character, not just a space.

[geforce]

Thanks alot

[DarrylBurke]

The 'word' character \\w is equivalent to [a-zA-Z_0-9] which allows digits and underscore in addition to alpha characters.

"Only letters allowed and at most one blank, which cannot be the first or last character"
What regular expression would meet this and why?

Assuming you're using matches() (not find())

"\\p{Alpha}+ {0,1}\\p{Alpha}+"

That literally says: one or more alpha characters, followed by zero or one space, followed by one or more alpha characters.

db