代码阻止应用程序打开?

时间:2014-01-23 13:38:49

标签: java android eclipse string split

我的代码是否阻止我的应用打开?没有错误,但它没有打开。或者是否存在导致冲突的拆分字符串的问题?

Mainactivity:

package com.pk.code;

import android.app.Activity;
import android.os.Bundle;
import android.view.Menu;
import android.widget.TextView;

public class MainActivity extends Activity {

    TextView smsText;@Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        smsText = (TextView) findViewById(R.id.smsText);

        Bundle extra = getIntent().getExtras();
        if (extra != null) {
            String SMSText = extra.get("sms").toString();
            String[] splitted = SMSText.split("-");
            if (splitted.length > 1 && splitted.length < 4) {
                smsText.setText(" " + splitted[0] + "," + " " + splitted[1]);
            } else {
                smsText.setText(SMSText);


            }
        }
    }


    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.main, menu);
        return true;
    }
}

1 个答案:

答案 0 :(得分:0)

很可能是extra.get(“sms”)中的问题,它返回null。检查你的logcat并在崩溃后在这里打印