Tom Wolve

Hi,
the String method 'indexOf(..)' is very helpful.
if it returns -1, the character is not in the string.

Hi,
if you're using a StringTokenizer with ' ' as delimiter character, use it with returnDelims enabled.
Count all tokens which are equal to " " and that should do it.

Hi,
try using 'new'.

Hi,
maybe you should use two interfaces, a 'basic' interface with one method and another one extending it where you may add the second method.
Greetings,
Tom

Hi,
try

Thread.sleep((int)(2000 + Math.random() * 2001));
and your thread should be sleeping between two and four seconds.

Hi,
to learn the basics of JDBC try the Sun tutorial.
JDBC tutorial

Hi,
let me give you some clues.
- 'nextToken' is a method. --> nextToken()
- String has no methods 'append' or 'insert'.
- The first character in a String is charAt(0).
This should make the it a little bit easier, at least for the compiler

Hi,
usually 'return' is the 'right' way to exit a method in Java, throwing an exception is another.

Hi,
take a look at JFace
JFaceDbc Homepage

Hi,
Let' try to make the code more readable to understand it.
.... i = 2; calc = i++; //-> calc is 2; i is 3; calc = calc + i++; //-> calc is 5, i is 4; i = calc; //-> i is 5
Hope this helps a bit
Tom

Hi,

char c = (char)Integer.parseInt("1100011",2);
This should do it.

Hi,
maybe it helps to know that unicodes like '\u000A' are replaced once before compiling.
So
//char a = '\u000A';
will be
//char a = ' ';
which will cause a compile error.
Greetings
Tom

Hi,
hope the following code fragment will help
..... String test = "ou=application"; StringTokenizer st = new StringTokenizer(test,"="); String ou = st.nextToken(); String application = st.nextToken(); // contains "application" ......

The one I use is a download.
If you think your one doesn't work, try to download again or try to re-install.

Follow the link or try search 'string equal'
Link