无法启动另一个“活动”在Android中 - 发生错误 [英] Can't start another "Activity" in Android - Gives Error

问题描述

这是一个较大代码的简化版本。我刚刚开始使用Android编程，并严重阻碍了这个问题一个多小时。

This is simplified version of a larger code. I've just started with Android programming and badly stuck with this problem for past an hour.

    /**Main Activity**/

    sumBut.setOnClickListener(new OnClickListener(){
        @Override
        public void onClick(View arg0) {
            Intent intent = new Intent(MainActivity.this, SumActivity.class);
            intent.putExtra("var1", et1.getText().toString());
            intent.putExtra("var2", et2.getText().toString());
            startActivity(intent);
        }
    });

此代码从文本框中获取两个变量，并将用户驱动到另一个活动，其中总和将显示数字。

This code takes two variable from the text boxes and drives the user to another activity where the sum of these numbers will be shown.

这是目标活动：

    public class SumActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_sum);

    TextView tv = (TextView) findViewById(R.id.textView1);
    Intent intent = getIntent();
    int a = Integer.parseInt(intent.getStringExtra("var1"));
    int b = Integer.parseInt(intent.getStringExtra("var2"));
    int c = a+b;
    tv.setText(c);
}

所有类的res / layout文件夹内都有关联的xml文件， ve创建。

There are associated xml files inside res/layout folder for all the classes I've created.

我的Manifest.xml文件中的示例

  <activity
        android:name="com.example.summer.MainActivity"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
    <activity android:name="com.example.summer.SumActivity"></activity>
    <activity android:name="com.example.summer.DifActivity"></activity>

LogCat

推荐答案

这是你的问题

 tv.setText(c);

将其更改为

 tv.setText(String.valueOf(c));

有不同的 setText（）方法您将在文档中看到这里。您调用的那个，它接受一个 int 作为参数，用于查找具有该'id'的资源，因此您需要将变量更改为字符串。这就是为什么你得到一个资源未找到异常。它不能找到 String资源与 id 的任何变量 c 是。

There are different setText() methods as you will see here in the docs. The one that you are calling, which accepts an int as a parameter, is used to find a resource with that 'id' so you need to change your variable to a String. This is why you get a Resource Not found exception. It can't find a String resource with the id of whatever variable c is.

只是一个旁注，但你可能想要将解析代码包裹在一个 try / catch 除非您正在进行其他错误检查，否则，如果用户输入非整数值，则可能会导致数字格式异常。

Just a side note but you might want to wrap your parsing code in a try/catch unless you are doing some other error checking or you may end up with a Number Format Exception if the user enters a non-integer value.

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