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[java] Should be an easy question

Started by Mulligan Oct 10, 2002 at 9:40 PM 1 reply 300+ views

Original Post

111

Oct 10, 2002 at 9:40 PM

I have the code: public void searchString( String str ) { String text = new String( textArea.getText() ); for( int a = 0; a < text.length() - str.length(); a++ ) { for( int b = 0; b < str.length(); b++ ) { if( text[a+b] != str[ b ] ) { break; } if( b == str.length() - 1 ) { System.out.println( "Found a match!!!" ); } } } } And I get the error: "Array required, but java.io.String found if( text[a+b] != str[ b ] )" I can't index a string? That seems strange if you ask me. Why can't I and how do I? [edited by - Mulligan on October 10, 2002 10:41:58 PM]

Martee

126

Martee

Oct 10, 2002 at 9:45 PM

You can use the String class''s charAt() method to get the character at a given position.

CaptainJester

101

CaptainJester

Oct 11, 2002 at 6:59 AM

First off, String is class in Java and classes cannot overload operators. So it is not like vextor in C++, where you can just index into it like an array.

So this line:

if( text[a+b] != str[ b ] )
{
break;
}

becomes(using charAt as Martee suggested):

if( text.charAt(a+b) != str.charAt( b ) )
{
break;
}

You should not do all this work, when the API does it for you. Just use:

int index=text.indexOf(str);

index will hold the starting position of str in text . It will hold -1 if str is not found.

Make it work.
Make it fast.

"I’m happy to share what I can, because I’m in it for the love of programming. The Ferraris are just gravy, honest!" --John Carmack: Forward to Graphics Programming Black Book

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