Question

我有以下示例，旨在捕捉String current = "d"不是double并且应该显示消息的事实。

当此代码运行时，由于NumberFormatException，编译器中会出现d。我相信我犯了一个非常简单的错误，但我只是看不到它。有人可以解释我哪里出错了吗？

public void runAddValidation()
{
    String current = "d";
    double vluRaised = Double.parseDouble(current);

    boolean vTypeDistanceBroke=false;

    try 
    { 
        double vFigure = vluRaised;
        vTypeDistanceBroke=false; 
    }

    catch (NumberFormatException nfe) 
    { 
        vTypeDistanceBroke=true; 
        System.out.println("Type-MoneyRaised: Failed"); 
        JOptionPane.showMessageDialog(null, "Please add a valid value (No Letters)");
    }
}

Answer 1

在parseDouble()

中插入try-catch

public void runAddValidation()
{
    double vluRaised = 0l;
    boolean vTypeDistanceBroke;
    String current = "d";
    double vFigure = vluRaised;

    try 
    { 
        vluRaised = Double.parseDouble(current);

        vTypeDistanceBroke=false; 
    }

    catch (NumberFormatException nfe) 
    { 
        vTypeDistanceBroke=true; 
        System.out.println("Type-MoneyRaised: Failed"); 
        JOptionPane.showMessageDialog(null, "Please add a valid value (No Letters)");
    }
}

根据评论中的建议：查看parseDouble() {{1}}，查看：

抛出：
   NullPointerException - 如果字符串为null
   NumberFormatException - 如果字符串不包含可解析的double。

Answer 2

您的Try / Catch不包含异常来源。代码会在它成功之前很久就会窒息。我也冒昧地解决了代码中的一些编译错误。你应该拥有的是：

   public void runAddValidation() {
        String current = "d";
        boolean vTypeDistanceBroke=false; 

        try{ 
           //This is where you'll have a problem
          double vluRaised = Double.parseDouble(current);
          double vFigure = vluRaised;

        }

        catch (NumberFormatException nfe){ 
               vTypeDistanceBroke=true; 
               System.out.println("Type-MoneyRaised: Failed"); 
               JOptionPane.showMessageDialog(null, "Please add a valid value(No Letters)");
       } 
    }

我试图捕获NumberFormatException但是异常仍在传递

2 个答案: