我正在尝试开发一个应用程序,它将从拨号盘中提取一个号码,然后呼叫另一个号码。但是我得到了一个号码格式异常,我想要呼叫的电话号码。
这就是代码。
public class testing extends BroadcastReceiver {
public void onReceive(Context context, Intent intent) {
if (intent.getAction().equals(Intent.ACTION_NEW_OUTGOING_CALL)) {
// get phone number from bundle
String phoneNumber = intent.getExtras().getString(Intent.EXTRA_PHONE_NUMBER);
// Toast.makeText(context, "Call: "+phoneNumber+" - ABORTING CALL.", Toast.LENGTH_LONG).show();
//
if (phoneNumber == "") {
phoneNumber = "0";
}
int checknumber = Integer.parseInt(phoneNumber);
if (checknumber == 4444) {
Toast.makeText(context, "hello", Toast.LENGTH_LONG).show();
setResultData(null);
Intent callIntent = new Intent(Intent.ACTION_CALL);
callIntent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
callIntent.setData(Uri.parse("tel:6768886877"));
context.startActivity(callIntent);
}
else {
Toast.makeText(context, "kkkk", Toast.LENGTH_LONG).show();
}
}
}
}下面是我的错误日志
01-04 20:16:24.941: E/AndroidRuntime(28536): java.lang.RuntimeException: Unable to start receiver vu.alhadji.vudial.testing: java.lang.NumberFormatException: Invalid int: "2156689451"
01-04 20:16:24.941: E/AndroidRuntime(28536): at android.app.ActivityThread.handleReceiver(ActivityThread.java:2126)
01-04 20:16:24.941: E/AndroidRuntime(28536): at android.app.ActivityThread.access$1500(ActivityThread.java:123)
01-04 20:16:24.941: E/AndroidRuntime(28536): at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1197)
01-04 20:16:24.941: E/AndroidRuntime(28536): at android.os.Handler.dispatchMessage(Handler.java:99)
01-04 20:16:24.941: E/AndroidRuntime(28536): at android.os.Looper.loop(Looper.java:137)
01-04 20:16:24.941: E/AndroidRuntime(28536): at android.app.ActivityThread.main(ActivityThread.java:4424)
01-04 20:16:24.941: E/AndroidRuntime(28536): at java.lang.reflect.Method.invokeNative(Native Method)
01-04 20:16:24.941: E/AndroidRuntime(28536): at java.lang.reflect.Method.invoke(Method.java:511)
01-04 20:16:24.941: E/AndroidRuntime(28536): at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:812)
01-04 20:16:24.941: E/AndroidRuntime(28536): at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:579)
01-04 20:16:24.941: E/AndroidRuntime(28536): at dalvik.system.NativeStart.main(Native Method)
01-04 20:16:24.941: E/AndroidRuntime(28536): Caused by: java.lang.NumberFormatException: Invalid int: "2156689451"
01-04 20:16:24.941: E/AndroidRuntime(28536): at java.lang.Integer.invalidInt(Integer.java:138)
01-04 20:16:24.941: E/AndroidRuntime(28536): at java.lang.Integer.parse(Integer.java:378)
01-04 20:16:24.941: E/AndroidRuntime(28536): at java.lang.Integer.parseInt(Integer.java:366)
01-04 20:16:24.941: E/AndroidRuntime(28536): at java.lang.Integer.parseInt(Integer.java:332)
01-04 20:16:24.941: E/AndroidRuntime(28536): at vu.alhadji.vudial.testing.onReceive(testing.java:24)
01-04 20:16:24.941: E/AndroidRuntime(28536): at android.app.ActivityThread.handleReceiver(ActivityThread.java:2119)
01-04 20:16:24.941: E/AndroidRuntime(28536): ... 10 more发布于 2013-01-05 09:38:46
这是因为对于int来说,6768886877这个数字太大了。
请改用long。
long checkNumber = 0l;
try{
checkNumber = Long.parseLong (phoneNumber);
}
catch (NumberFormatException e)
{
e.printStackTrace();
}发布于 2013-01-05 09:42:09
对于整数来说,2156689451太大了(范围从-2^31≈-2,147E9到+2^31-1≈2,147E9,而您的数字是~2,15E10)
使用long代替int...
https://stackoverflow.com/questions/14167586
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