Question

我正在尝试编写一个try catch块，如果用户没有输入名称而只是按enter或okay我将会抛出异常，因为它没有抛出任何它只是接受空白并继续。有人能帮我吗？提前谢谢！

这是功能：

public String setOwnerName() {
        boolean isName = false;

        while(!isName) {
            try {
                this.ownerName = JOptionPane.showInputDialog(null, "Enter the account owner's name.", "Owner's Name", JOptionPane.PLAIN_MESSAGE);
                if(this.ownerName != "") {
                    isName = true;  
                }
            } catch(IllegalArgumentException e) {
                JOptionPane.showMessageDialog(null, "Error you did not enter a name, please try again.", "Error", JOptionPane.ERROR_MESSAGE);
            }
        }
        return this.ownerName;
    }

Answer 1

空名称不会exception。你需要手动检查它。尝试，

public String setOwnerName() {
  boolean isName = false;

while(!isName) {               
 ownerName = JOptionPane.showInputDialog(null, "Enter the account owner's name.",
               "Owner's Name", JOptionPane.PLAIN_MESSAGE);

  if(ownerName.trim().isEmpty()){
    JOptionPane.showMessageDialog(null,
        "Error you did not enter a name, please try again.", 
       "Error", JOptionPane.ERROR_MESSAGE);
  }
  else{
      isName = true;
      }               
   }// end of while

    return ownerName;
}

Answer 2

这样做

 try{           
      if(ownerName.trim().isEmpty()) 
      {
          throw new IllegalArgumentException("Input is wrong");
      }
 } catch(IllegalArgumentException e) {

 }

如果JTextField为空，则抛出错误

2 个答案: