another simple Java question: string search

[HardTech]

Say I have two strings that say "American Dollars" and "Australian Dollars"

I make another string that says "Dollar"... is there a way I can compare "Dollar" to the two strings above and have it return a true?

is there a partial match method in the String class?

 

[MrChad]

You could use indexOf() or matches() (if you want a regular expression search).

 

[HardTech]

sweet... I would have never guessed that looking at Sun's website.

Thanks a bunch!!

 

[notfred]

Originally posted by: HardTech
sweet... I would have never guessed that looking at Sun's website.

Thanks a bunch!!

Why not? They're right there in the list of methods.

 

[HardTech]

hmm.. it's not working

so let's say:

String USD = "American Dollars";
String AUD = "Australian Dollars";
String HardTech = "Dollars";

HardTech.indexOf(USD) would return a value >= 1?

 

[HardTech]

Originally posted by: notfred

Originally posted by: HardTech
sweet... I would have never guessed that looking at Sun's website.

Thanks a bunch!!

Why not? They're right there in the list of methods.

I guess I never translated "Returns the index within this string of the first occurrence of the specified substring." into "contains"

 

[notfred]

Originally posted by: HardTech
hmm.. it's not working

so let's say:

String USD = "American Dollars";
String AUD = "Australian Dollars";
String HardTech = "Dollars";

HardTech.indexOf(USD) would return a value >= 1?

you want: USD.indexOf(HardTech);

 

[MrChad]

Originally posted by: HardTech
hmm.. it's not working

so let's say:

String USD = "American Dollars";
String AUD = "Australian Dollars";
String HardTech = "Dollars";

HardTech.indexOf(USD) would return a value >= 1?

No, but USD.indexOf(HardTech) should. Read the documentation for that method again.

 

[znaps]

You can also say

USA.endsWith(HardTech)

Which would return a boolean value of true.

P.S don't name local variables starting with a capital letter, it's breaks the conventional approach which would be hardTech.