如何在android中打破数组字符串文本？

Question

你好！！ 我想这样显示表格：-

• 1 * 1 = 1
• 1 * 2 = 2
• -

• 1 * 10 = 10

  但我可以显示第一行然后它消失然后显示下一行。我正在尝试使用字符串数组来显示数字。

这是我的代码：-

public class MainActivity extends Activity {
    TextView txtView; 
    static String test[]={"","1","","1 *","","1 * 1","","1 * 1 = 1",
        "","1","","1 *","","1 * 2","","1 * 2 = 2",
        "","1","","1 *","","1 * 3","","1 * 3 = 3",
        "","1","","1 *","","1 * 4","","1 * 4 = 4",
        "","1","","1 *","","1 * 5","","1 * 5 = 5",
        "","1","","1 *","","1 * 6","","1 * 6 = 6",
        "","1","","1 *","","1 * 7","","1 * 7 = 7",
        "","1","","1 *","","1 * 8","","1 * 8 = 8",
        "","1","","1 *","","1 * 9","","1 * 9 = 9",
        "","1","","1 *","","1 * 10","","1 * 10 = 10"};
    Handler handler;
    ImageView imageView;
    Runnable runnable;
    StringBuilder sb = new StringBuilder();



    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        Intent i = getIntent(); 
        String someVariable = i.getStringExtra("MA");
        Toast.makeText(getApplicationContext(), "SomeVariable" +someVariable, 222).show();

        txtView = (TextView) findViewById(R.id.greentxt_name);



     handler = new Handler();
    runnable = new Runnable() {

      int t=0;
      public void run() {

          sb.append(test[t]);

          t++;

          if( t>=79)
          { 


           handler.removeCallbacks(runnable) ;
           txtView.setText(test[t]);

          }
          else
          { txtView.setText(test[t]);
          handler.postDelayed(this,1000);    
          }
      }
  };
  handler.postDelayed(runnable, -1000); //for initial delay..
}


}

任何答案都是可观的。

提前致谢

Answer 1

你可以这样做。 ：

StringBuffer str= new StringBuffer();
for(int i=1;i<=10;i++)
{
    str.append(""+1+" * "+i+" = "+(1*i)+"\n");
    Log.e("", ""+str.toString());
}
txt.setText(str);

Answer 2

我已经对您的代码进行了一些编辑以达到预期的效果 就是这样：

    public class MainActivity extends Activity {

    TextView txtView; 
    static String test[]={"1"," * ","1"," = ","1", "\n",
        "1"," * ","2"," = ","2", "\n",
        "1"," * ","3"," = ","3", "\n",
        "1"," * ","4"," = ","4", "\n",
        "1"," * ","5"," = ","5", "\n",
        "1"," * ","6"," = ","6", "\n",
        "1"," * ","7"," = ","7", "\n",
        "1"," * ","8"," = ","8", "\n",
        "1"," * ","9"," = ","9", "\n",
        "1"," * ","10"," = ","10"};

    Handler handler;
    ImageView imageView;
    Runnable runnable;
    StringBuilder sb = new StringBuilder();

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

        Intent i = getIntent(); 
        txtView = (TextView) findViewById(R.id.greentxt_name);



     handler = new Handler();
     runnable = new Runnable() {

      int t=0;
      public void run() {

          sb.append(test[t]);

          t++;

          if( t>=59)
          { 


           handler.removeCallbacks(runnable) ;
           txtView.setText(sb);

          }
          else
          { txtView.setText(sb);
          handler.postDelayed(this,1000);    
          }

        }};
     handler.postDelayed(runnable, -1000); //for initial delay.. 
}}

您可以删除硬编码的值并对其进行概括，以便将来用于任何其他表。

Answer 3

文本视图中的文本不断被替换，因为来自 text[t] 的新索引值

您可以通过将行连接为 sb 字符串中的下一行并替换来稍微更改逻辑

txtView.setText(test[t]); 

由

txtView.setText(sb);