伙计们sms manager关于短信计划的问题...我在使用的清单中使用了
`<uses-permission android:name="android.permisson.SEND_SMS"/>`
但我收到了这个错误...我不知道为什么......我告诉你我的LOGCAT：
FATAL EXCEPTION: main
Process: com.example.alarm, PID: 31456
java.lang.RuntimeException: Unable to start receiver com.example.alarm.AlarmReceiver: java.lang.SecurityException: Sending SMS message: uid 10051 does not have android.permission.SEND_SMS.
at android.app.ActivityThread.handleReceiver(ActivityThread.java:2615)
at android.app.ActivityThread.access$1700(ActivityThread.java:151)
at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1397)
at android.os.Handler.dispatchMessage(Handler.java:110)
at android.os.Looper.loop(Looper.java:193)
at android.app.ActivityThread.main(ActivityThread.java:5327)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:515)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:824)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:640)
at dalvik.system.NativeStart.main(Native Method)
Caused by: java.lang.SecurityException: Sending SMS message: uid 10051 does not have android.permission.SEND_SMS.
at android.os.Parcel.readException(Parcel.java:1465)
at android.os.Parcel.readException(Parcel.java:1419)
at com.android.internal.telephony.ISms$Stub$Proxy.sendText(ISms.java:926)
at android.telephony.SmsManager.sendTextMessage(SmsManager.java:156)
at com.example.alarm.AlarmReceiver.onReceive(AlarmReceiver.java:24)
at android.app.ActivityThread.handleReceiver(ActivityThread.java:2597)
at android.app.ActivityThread.access$1700(ActivityThread.java:151)
at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1397)
at android.os.Handler.dispatchMessage(Handler.java:110)
at android.os.Looper.loop(Looper.java:193)
at android.app.ActivityThread.main(ActivityThread.java:5327)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:515)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:824)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:640)
at dalvik.system.NativeStart.main(Native Method)
关于接收方广播的代码：
public class AlarmReceiver extends BroadcastReceiver{
@Override
public void onReceive(Context context, Intent intent) {
// TODO Auto-generated method stub
String phoneNumberReciver="123456";
String message="blablabla";
/*String SPhone =i.getStringExtra("exPhone");
String SSms = i.getStringExtra("exSmS");*/
//android.telephony.SmsManager sms= SmsManager.getDefault();
//sms.sendTextMessage(phoneNumberReciver, null, message, null, null);
SmsManager smsManager = SmsManager.getDefault();
smsManager.sendTextMessage(phoneNumberReciver, null, message, null, null);
Toast.makeText(context, "Alarm Triggered and SMS Sent", Toast.LENGTH_LONG);
}
}
这是我的表现：
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.alarm"
android:versionCode="1"
android:versionName="1.0" >
<uses-sdk
android:minSdkVersion="8"
android:targetSdkVersion="17" />
<uses-permission android:name="com.example.alarm.permission.SET_ALARM"/>
<uses-permission android:name="android.permisson.SEND_SMS"/>
<uses-permission android:name="android.permission.READ_CONTACTS"/>
<application
android:allowBackup="true"
android:icon="@drawable/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme" >
<activity
android:name="com.example.alarm.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>
<receiver android:name=".AlarmReceiver"/>
</application>
</manifest>
世界卫生组织可以帮助我吗？大家先谢谢大家！
答案 0 :(得分：3)
复制并粘贴此代码，以便发送短信服务。
Button sendBtn = (Button)findViewById(R.id.senbtn);
sendBtn.setOnClickListener(new View.OnClickListener() {
public void onClick(View view) {
sendSMSMessage();
}
});
}
protected void sendSMSMessage() {
phoneNo = txtphoneNo.getText().toString();
message = txtMessage.getText().toString();
if (ContextCompat.checkSelfPermission(this,
Manifest.permission.SEND_SMS)
!= PackageManager.PERMISSION_GRANTED) {
if (ActivityCompat.shouldShowRequestPermissionRationale(this,
Manifest.permission.SEND_SMS)) {
} else {
ActivityCompat.requestPermissions(this,
new String[]{Manifest.permission.SEND_SMS},
MY_PERMISSIONS_REQUEST_SEND_SMS);
}
}
}
@Override
public void onRequestPermissionsResult(int requestCode,String permissions[], int[] grantResults) {
switch (requestCode) {
case MY_PERMISSIONS_REQUEST_SEND_SMS: {
if (grantResults.length > 0
&& grantResults[0] == PackageManager.PERMISSION_GRANTED) {
SmsManager smsManager = SmsManager.getDefault();
smsManager.sendTextMessage(phoneNo, null, message, null, null);
Toast.makeText(getApplicationContext(), "SMS sent.",
Toast.LENGTH_LONG).show();
} else {
Toast.makeText(getApplicationContext(),
"SMS faild, please try again.", Toast.LENGTH_LONG).show();
return;
}
}
}
}
在menifest文件中添加：
<uses-permission android:name="android.permission.SEND_SMS"/>
答案 1 :(得分：2)
第1步 您的明确许可是
<uses-permission android:name="android.permisson.SEND_SMS"/>
用
更新它<uses-permission android:name="android.permission.SEND_SMS"/>
其
android.permission.SEND_SMS不是android.permisson.SEND_SMS
权限拼写错误
<强>步骤2
<receiver android:name=".AlarmReceiver">
<intent-filter>
<action android:name="android.provider.Telephony.SMS_RECEIVED" />
</intent-filter>
</receiver>
答案 2 :(得分：0)
您需要为OS marshmallow及以上
提供运行时权限完成此link
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
//SEE NEXT STEP
}else{
sendSms();
}
if (checkSelfPermission(Manifest.permission.SEND_SMS) != PackageManager.PERMISSION_GRANTED) {
//SEE NEXT STEP
} else {
sendSMS();
}
if (shouldShowRequestPermissionRationale(Manifest.permission.SEND_SMS)) {
//show some description about this permission to the user as a dialog, toast or in Snackbar
} else {
//request for the permission
}
@Override
public void onRequestPermissionsResult(int requestCode, String[] permissions, int[] grantResults) {
super.onRequestPermissionsResult(requestCode, permissions, grantResults);
if (grantResults[0] == PackageManager.PERMISSION_GRANTED) {
Snackbar.make(findViewById(R.id.rl), "Permission Granted",
Snackbar.LENGTH_LONG).show();
sendSMS();
} else {
Snackbar.make(findViewById(R.id.rl), "Permission denied",
Snackbar.LENGTH_LONG).show();
}
}
答案 3 :(得分：0)
经过数小时的调试，我发现将此权限添加到xml文件中不再产生错误：
<uses-permission-sdk-23 android:name="android.permission.READ_PHONE_STATE"/>
答案 4 :(得分：0)
当前未获得许可的主要原因是因为您的项目的targetSdkVersion为23或更高，并且您所请求的许可是“危险的”。在Android 6.0中 您可以手动授予权限：
const onPermissionHandle = async () => {
// We need to ask permission for Android only
if (Platform.OS === 'android') {
// Calling the permission function
alert(granted);
const granted = await PermissionsAndroid.request(
PermissionsAndroid.PERMISSIONS.SEND_SMS,
{
title: 'Example App SEND_SMS Permission',
message: 'Example App needs access to your SEND_SMS',
},
);
alert(granted);
if (granted === PermissionsAndroid.RESULTS.GRANTED) {
// Permission Granted
alert('You can use the SEND_SMS');
} else {
// Permission Denied
alert('SEND_SMS Permission Denied');
}
} else {
alert('You can use the SEND_SMS ');
}
};