asked on

e.getSource()

Hi Experts,

What did I do wrong in the following line 01 ? and how do I fix it ?
where I have :
b1 = new JButton("1"); b1.addActionListener(this); Pane2.add(b1) ;
b2 = new JButton("2"); b2.addActionListener(this); Pane2.add(b2) ;
------------------------------------------------------------------
public void actionPerformed(ActionEvent e) {
n++ ;
switch((int)e.getSource()){ //line 01 inconvertible types, found : java.Lang.Object, require : int
case '1' : myString += '1'; break ;
case '2' : myString += '2'; break ;
}

----------------------------------------------------------------------
please help !

ASKER CERTIFIED SOLUTION

lhankins

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meow00

ASKER

so .... does this mean : I must use

if(e.getSource() == b1) {} ... I can't use switch ???
that seems to be complicated to use a lot of if & else .... is there a better way to use switch here ?

Thanks !

expertmb

it returns the object , which is a component.
instead use
b1.setName("button1");
b2.setName("button2");
if(e.getSource() instanceof JButton) {
if(((JButton)e.getSource()).getName() == "button1")
dosomething

Javatm

Where is the full code ? and what's the main problem ?

SOLUTION

expertmb

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armoghan

if(((JButton)e.getSource()).getText() == "1")
myString +=1;
if(((JButton)e.getSource()).getText() == "2")
myString +=2;

sudhakar_koundinya

public void actionPerformed(ActionEvent e) {
n++ ;
switch((int)e.getSource().getText().toString().charAt(0)){
case '1' : myString += '1'; break ;
case '2' : myString += '2'; break ;
}

SOLUTION

armoghan

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SOLUTION

sudhakar_koundinya

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SOLUTION

expertmb

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sudhakar_koundinya

or you may try like this

public void actionPerformed(ActionEvent e) {
n++ ;
switch(e.getSource().getName().toString().charAt(0)){
case '1' : myString += '1'; break ;
case '2' : myString += '2'; break ;
}

SOLUTION

talvio

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sudhakar_koundinya