android 程序崩溃时，没有输入输入和按钮被按下

jk9hmnmh 于 2023-11-15 发布在 Android

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你好，我正在做一个随机数字猜测游戏。问题是，如果用户没有输入任何数字，并按下按钮，如果条件的answerEmpty变量应执行，并在结果文本中显示“请输入一个数字”。但程序崩溃时，我按下按钮。else if和else条件工作，但启动if条件不工作。请帮助

class MainActivity : AppCompatActivity() {
        override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
        setContentView(R.layout.activity_main)
        //Calling button method
        val button = findViewById<Button>(R.id.guessBtn)
        //Set on click listener for the button
        button.setOnClickListener {
            buttonPressed()
           }
        }
       private fun buttonPressed() {
            //Method to find view from activity_main
            val result = findViewById<TextView>(R.id.result)
            val userAnsEditText = findViewById<EditText>(R.id.answer)
            val userAnsText = userAnsEditText.text.toString()
            val userAnsInt = userAnsText.toInt()
            //This variable will display the answer
            val answerCorrect = "Congrats, you have correctly guessed the number"
            val answerWrong = "Sorry,you have guessed the wrong number"
            val answerEmpty = "Please enter a number"
            //Making random numbers from 1 to 10
            val randomValues = Random.nextInt(1, 11)
            //Adding a condition to check if user answer matches the random answer
            if(userAnsInt == null){
                result.text = answerEmpty
            }
            else if(userAnsInt == randomValues){
                result.text = answerCorrect
            }
            else{
                result.text = answerWrong
            }

          }

字符串

android

来源：https://stackoverflow.com/questions/77457827/program-crashes-when-no-input-is-entered-and-button-is-pressed

2条答案

按热度按时间

1cklez4t1#

在int中解析之前，应检查EditText文本是否为空或非条件
添加以下条件

if (userAnsText.isEmpty()) {
    result.text = answerEmpty
}

字符串
val userAnsInt = userAnsText.toInt()线之前。
它会修复你的崩溃。
添加isEmpty()条件后，buttonPressed()函数如下所示。

private fun buttonPressed() {
    //Method to find view from activity_main
    val result = findViewById<TextView>(R.id.result)
    val userAnsEditText = findViewById<EditText>(R.id.answer)
    val userAnsText = userAnsEditText.text.toString()

    val answerCorrect = "Congrats, you have correctly guessed the number"
    val answerWrong = "Sorry,you have guessed the wrong number"
    val answerEmpty = "Please enter a number"

    if (userAnsText.isEmpty()) {
        result.text = answerEmpty
    } else {
        val userAnsInt = userAnsText.toInt()
        //This variable will display the answer
        //Making random numbers from 1 to 10
        val randomValues = Random.nextInt(1, 11)
        //Adding a condition to check if user answer matches the random answer
        //when condition more relavant in kotlin than if-else if-else
        when (userAnsInt) {
            null -> result.text = answerEmpty
            randomValues -> result.text = answerCorrect
            else -> result.text = answerWrong
        }
    }
}

型

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kpbwa7wx2#

例如，使用Try catch块

`

try {
    // Code that may cause an exception
    int result = 10 / 0; // This will throw an ArithmeticException
    Log.d("MyApp", "Result: " + result); // This line won't be executed
} catch (ArithmeticException e) {
    // Handle the exception
    Log.e("MyApp", "An arithmetic exception occurred: " + e.getMessage());
}

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“下面是各部分的细目：

**try：**可能引发异常的代码放在此块中。**catch：**如果try块中出现异常，则执行catch块。它指定要捕获的异常类型和处理它的代码。**finally：**此块包含无论是否出现异常都会执行的代码。它是可选的。之后您的应用不会崩溃。它会显示错误，您可以在吐司或输出窗口中打印错误。

赞(0）回复(0）举报 2023-11-15

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android 程序崩溃时，没有输入输入和按钮被按下

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