java - 为什么我的应用程序在我单击某个按钮时崩溃？

Question

大家好，我是 Java 的新手，下面的代码肯定会证明这一点。对于我哪里出错的任何提示，我将不胜感激，谢谢!

我正在尝试创建一个简单的应用程序来显示随机引语，并提供猜出这句话的人的机会。

很确定主要问题与 ---> if (guessNameInt == whoSaidItInt)因为它只有在我单击启用 tryLuck if 语句的按钮时才会崩溃。

下面是我的代码

int randomNum;
String whoSaidIt;

// quotes and numbers
public void randomRick(View view) {
    if (randomNum == 0) {

        Toast.makeText(getApplicationContext(), "Life is effort and I'll stop when I die!", Toast.LENGTH_LONG).show();
        whoSaidIt = "Jerry";
    }

    if (randomNum == 1) {

        Toast.makeText(getApplicationContext(), "Well look where being smart got you.", Toast.LENGTH_LONG).show();
        whoSaidIt = "Jerry";
    }

    if (randomNum == 2) {

        Toast.makeText(getApplicationContext(), "Ohh yea, you gotta get schwifty.", Toast.LENGTH_LONG).show();
        whoSaidIt = "Rick";

    }

}

public void tryLuck(View view) {
    EditText guessedName = (EditText) findViewById(R.id.authorIs);
    String guessedNameString = guessedName.getText().toString();
    int guessNameInt = Integer.parseInt(guessedNameString);
    int whoSaidItInt = Integer.parseInt(whoSaidIt);

    if (guessNameInt == whoSaidItInt) {

        Toast.makeText(getApplicationContext(), "Holy crow, Good job!", Toast.LENGTH_LONG).show();

    }

    else {
        Toast.makeText(getApplicationContext(), "Try again", Toast.LENGTH_LONG).show();
    }

}

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    Random randomGenerator = new Random();

    randomNum = randomGenerator.nextInt(3);
}`

Answer 1

您正在尝试将 String 转换为 Int。这仅在字符串是实际 int(即数字)而不是名称(Jerry 或 Rick)时有效。请改用 guessNameString.equals(whoSaidIt)。