java - 为什么我仍然收到 NumberFormatException 错误？

Question

好吧，我很确定我已经处理了这个错误，但是每当我尝试测试一个字符串代替整数时，它都会不断发送错误，尽管我指定了处理它。

我使文本字段只接受整数，但是我希望它显示，如果输入字符串，它将发送一个 JOptionPane 说只允许数字。

这是我的整个代码:

    public void giveItem() {
    try {
        String itemId = textField1.getText();
        String itemAmount = textField2.getText();

    if (isTargetEmpty()){
        JOptionPane.showMessageDialog(
                null, "Please enter a player to continue!", "Error", JOptionPane.INFORMATION_MESSAGE
        );
    } else if (itemId.isEmpty() || itemAmount.isEmpty()){
        JOptionPane.showMessageDialog(
                null,"Please fill the required fields!", "Error",JOptionPane.INFORMATION_MESSAGE
        );
    } else {

        String player = getTargetName();
        Player target = World.getPlayerByDisplayName(player);
        if (target != null) {

            int id = Integer.parseInt(itemId);
            int amount = Integer.parseInt(itemAmount);

            if (Double.isNaN(id) || Double.isNaN(amount)){
                JOptionPane.showMessageDialog(
                        null, "You can only enter numbers!", "Error!", JOptionPane.INFORMATION_MESSAGE
                );

        } else if (amount <= target.getInventory().getFreeSlots()) {

            target.getInventory().addItem(id, amount);
            target.getPackets().sendGameMessage(
                "You have received " + (amount > 1 ? " Items!" : "an Item!")
        );
        JOptionPane.showMessageDialog(
                null, Utils.formatPlayerNameForDisplay(target.getUsername()) + " has received the " +
                (amount > 1 ? "Items successfully in his/her inventory." : "Item successfully in his/her inventory.")
        );
        logAction(SPSutil.getTime() + Utils.formatPlayerNameForDisplay(target.getUsername())
                + " has received the item : "  + id + ". Amount : " + amount + " in his/her inventory."
        );
        } else {
            target.getBank().addItem(id, amount, true);
            target.getPackets().sendGameMessage(
                    "You have received " + (amount > 1 ? "Items in your bank!" : "an Item in your bank!")
            );
            JOptionPane.showMessageDialog(
                    null, Utils.formatPlayerNameForDisplay(target.getUsername()) + " has received the " +
                    (amount > 1 ? "Items successfully in his/her bank." : "Item successfully in his/her bank.")
            );
            logAction(SPSutil.getTime() + Utils.formatPlayerNameForDisplay(target.getUsername())
                    + " has received the item : "  + id + ". Amount : " + amount + " in his/her bank."
            );
        }
    } else {
            JOptionPane.showMessageDialog(
                    null , "Player does not exist!", "Error" , JOptionPane.INFORMATION_MESSAGE
            );
        }
 }
    } catch(IOException e){
        e.printStackTrace();
    }
}

Answer 1

Double.isNaN()解析后将不会检查数字是否有效。此方法的参数是一个 double(因此您的 id 会自动提升)，并且它仅检查此 double 是否为 Double.NaN 。 int 永远无法做到这一点，甚至无法提升。

你必须做的是:

try {
    int id = Integer.parseInt(itemId);
} catch (NumberFormatException e) {
    // not an integer
}