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我的 android 应用程序在收到消息时首先显示一个 toast,当它收到来自特定号码的消息时,它会显示另一个 toast。但它没有显示第二个吐司。

这是我的代码:

import android.content.BroadcastReceiver;
import android.content.Context;
import android.content.Intent;
import android.os.Bundle;
import android.telephony.SmsManager;
import android.telephony.SmsMessage;
import android.util.Log;
import android.widget.Toast;


public class IncomingSms extends BroadcastReceiver {

    // Get the object of SmsManager
    final SmsManager sms = SmsManager.getDefault();

    public void onReceive(Context context, Intent intent) {

        // Retrieves a map of extended data from the intent.
        final Bundle bundle = intent.getExtras();

        try {

            if (bundle != null) {

                final Object[] pdusObj = (Object[]) bundle.get("pdus");

                for (int i = 0; i < pdusObj.length; i++) {

                    SmsMessage currentMessage = SmsMessage.createFromPdu((byte[]) pdusObj[i]);
                    String phoneNumber = currentMessage.getDisplayOriginatingAddress();

                    String senderNum = phoneNumber;
                    String message = currentMessage.getDisplayMessageBody();

                    Log.i("SmsReceiver", "senderNum: "+ senderNum + "; message: " + message);

                    int duration = Toast.LENGTH_SHORT;
                    Toast toast = Toast.makeText(context, "senderNum: "+ senderNum + ", message: " + message, duration);
                    toast.show();
                    String serverNumber= "+919886096376";
                    if(senderNum == serverNumber)
                    {
                        Toast toast1 = Toast.makeText(context,"alert message received!!!!",Toast.LENGTH_LONG);
                        toast.show();
                    }

                } // end for loop
              } // bundle is null

        } catch (Exception e) {
            Log.e("SmsReceiver", "Exception smsReceiver" +e);

        }
    }



}
4

2 回答 2

3

if(senderNum == serverNumber)以错误的方式进行比较。在 Java 和 Android==中用于比较objects不是字符串。当你想比较两个字符串时,你需要使用.equals()方法。

你需要像下面这样比较,

if(senderNum.equals(serverNumber))
{
      Toast toast1 = Toast.makeText(context,"alert message received!!!!",Toast.LENGTH_LONG);
      toast1.show();     // change this to toast1
}
于 2014-02-25T10:44:36.223 回答
1

改变这个

if(senderNum == serverNumber){
    Toast toast1 = Toast.makeText(context,"alert message received!!!!",Toast.LENGTH_LONG);
    toast.show();
}

对此

if(senderNum.equals(serverNumber)){
    Toast toast1 = Toast.makeText(context,"alert message received!!!!",Toast.LENGTH_LONG);
    toast1.show();
}

你没有在正确的吐司上打电话。

此外,比较 2 个字符串的正确方法如下:

senderNum.equals(serverNumber)

通过使用该==运算符,您正在比较对 String 对象的引用,因此它很可能会返回 false。

于 2014-02-25T10:44:51.733 回答