Java JOptionPane,返回时的帮助
这是我在这个网站上的第一篇帖子,因为这个网站已经帮了我很多忙。我想到的不是什么问题,而是一些我想学习如何去做的事情。这是我的密码Java JOptionPane,返回时的帮助,java,string,joptionpane,parseint,Java,String,Joptionpane,Parseint,这是我在这个网站上的第一篇帖子,因为这个网站已经帮了我很多忙。我想到的不是什么问题,而是一些我想学习如何去做的事情。这是我的密码 String a = JOptionPane.showInputDialog(null,"Please pick something for me to do master:\nMynumber,Read2me, Conversions"); if (a.equals("Mynumber")) { MyNumber = Integer.parseInt(JOp
String a = JOptionPane.showInputDialog(null,"Please pick something for me to do master:\nMynumber,Read2me, Conversions");
if (a.equals("Mynumber"))
{
MyNumber = Integer.parseInt(JOptionPane.showInputDialog(null, "Please enter your number: "));
String b = JOptionPane.showInputDialog(null,"Was Your number "+MyNumber+"\n Y/N");
if (b.equals("y"))
{
JOptionPane.showMessageDialog(null,"Good...good, now lets play with\n your number");
}
else if(b.equals("N"))
{
JOptionPane.showMessageDialog(null,"returning");
我的目标是,当它到达最后一个on(当用户键入n)时,我希望它返回到起始字符串或字符串A,例如,如何在代码中实现这一点
while(true) { // instead of while(true) you can also write other condition
String a = JOptionPane.showInputDialog(null,"Please pick something for me to do master:\nMynumber,Read2me, Conversions");
if (a.equals("Mynumber")) {
MyNumber = Integer.parseInt(JOptionPane.showInputDialog(null, "Please enter your number: "));
String b = JOptionPane.showInputDialog(null,"Was Your number "+MyNumber+"\n Y/N");
if (b.equals("y")) {
JOptionPane.showMessageDialog(null,"Good...good, now lets play with\n your number");
}
else if(b.equals("N")) {
JOptionPane.showMessageDialog(null,"returning");
}
}
}
他还需要一个返回语句和适当的方法签名。你到底想要什么?返回哪里