Java 空文本崩溃

我刚开始学习Android，我编写了一个代码，给你一个等式，你需要解它，这很简单，但如果有人按下检查按钮，编辑文本为空，应用程序就会崩溃，我希望它算作错误。。这是我的密码：

   public void Generate(View v) {
    x1=10+(int)((99-10+1)*Math.random());
    x2=10+(int)((99-10+1)*Math.random());
    tv1.setText("" + x1 + "+" + x2 + "=" + "?");
}

public void Check(View view) {
    answer1 = et1.getText().toString();
    answer = Integer.parseInt(answer1);
    if (answer == (x1 + x2)) {
        count1++;
        count2++;
        Toast.makeText(getApplicationContext(), "Correct !",
                Toast.LENGTH_SHORT).show();
    }


    if(answer!=(x1+x2))
    {
        count1++;
        count3++;
        Toast.makeText(getApplicationContext(), "Wrong !",
                Toast.LENGTH_SHORT).show();
    }

    if(answer1=="")
    {
        count1++;
        count3++;
        Toast.makeText(getApplicationContext(), "Don't try to cheat !",
                Toast.LENGTH_SHORT).show();
    }

    et1.setText("");
    tv2.setText("Number of questions : "+count1);
    tv3.setText("Right answers : "+count2);
    tv4.setText("Wrong answers : "+count3);

    Generate(view);
}

---

问题是，您正在将et1.getText（）转换为字符串，如果返回null值，则会引发null指针异常。所以需要执行一个检查值是否为空：

if(et1.getText()!=null){

answer1 = et1.getText().toString();
if(et1.getText().toString().length()>0){
    answer = Integer.parseInt(answer1);
}
    if (answer == (x1 + x2)) {
        count1++;
        count2++;
        Toast.makeText(getApplicationContext(), "Correct !",
                Toast.LENGTH_SHORT).show();
    }


    if(answer!=(x1+x2))
    {
        count1++;
        count3++;
        Toast.makeText(getApplicationContext(), "Wrong !",
                Toast.LENGTH_SHORT).show();
    }

    if(answer1=="")
    {
        count1++;
        count3++;
        Toast.makeText(getApplicationContext(), "Don't try to cheat !",
                Toast.LENGTH_SHORT).show();
    }

    et1.setText("");
    tv2.setText("Number of questions : "+count1);
    tv3.setText("Right answers : "+count2);
    tv4.setText("Wrong answers : "+count3);

    Generate(view);
}

另一件事是，若字符串是空的，那个么它将无法解析为int，所以您还需要对此进行检查

或者

if(et1.getText().toString().length()>0){
        answer = Integer.parseInt(answer1);
    }

或

---

您可以检查EditText是否为空。在你的代码中，按照下面的方式执行

if (!et1.getText().toString().matches(""))
{
    answer1 = et1.getText().toString();
}
else
{
    //No value entered
}

或者干脆

if (!et1.getText().toString().equals(""))
{
    answer1 = et1.getText().toString();
}
else
{
    //No value entered
}

---

您遇到了什么异常？在尝试

answer=Integer.parseInt（answer1）

之前，请检查字符串是否为空。要检查字符串是否为空，请使用

TextUtils.isEmpty（answer1）

NumberFormatException-如果字符串不包含可解析的int.

I betIt可以是任何内容，则必须使用android studio logger并发布实际堆栈跟踪。你的问题毫无意义，它可以是任何东西。提供更多细节

if (!et1.getText().toString().equals(""))
{
    answer1 = et1.getText().toString();
}
else
{
    //No value entered
}