所以,在过去的两天里,我在尝试连接时遇到了这个错误。
尝试了 UDP、TCP 和 Httprequest/Httppost。
我们尝试在两台设备上使用 APK
尝试使用 WIFI,我们添加了 3 个权限,尽管据我所知,我只需要互联网权限。添加权限而不是错误。
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.INTERACT_ACROSS_USERS_FULL" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
05-21 06:21:28.809: E/DatabaseUtils(2347): Writing exception to parcel
05-21 06:21:28.809: E/DatabaseUtils(2347): java.lang.SecurityException: Permission Denial: get/set setting for user asks to run as user -2 but is calling from user 0; this requires android.permission.INTERACT_ACROSS_USERS_FULL
05-21 06:21:28.809: E/DatabaseUtils(2347): at com.android.server.am.ActivityManagerService.handleIncomingUser(ActivityManagerService.java:13140)
05-21 06:21:28.809: E/DatabaseUtils(2347): at android.app.ActivityManager.handleIncomingUser(ActivityManager.java:2038)
05-21 06:21:28.809: E/DatabaseUtils(2347): at com.android.providers.settings.SettingsProvider.callFromPackage(SettingsProvider.java:607)
05-21 06:21:28.809: E/DatabaseUtils(2347): at android.content.ContentProvider$Transport.call(ContentProvider.java:279)
05-21 06:21:28.809: E/DatabaseUtils(2347): at android.content.ContentProviderNative.onTransact(ContentProviderNative.java:273)
05-21 06:21:28.809: E/DatabaseUtils(2347): at android.os.Binder.execTransact(Binder.java:388)
05-21 06:21:28.809: E/DatabaseUtils(2347): at dalvik.system.NativeStart.run(Native Method)
编辑:我还添加了“android:sharedUserId="android.uid.system"”,这是完全相同的错误……仍然……
编辑:在此处添加 list :
<?xml version="1.0" encoding="utf-8"?>
<application
android:allowBackup="true"
android:icon="@mipmap/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme"
android:sharedUserId="android.uid.system">
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.INTERACT_ACROSS_USERS_FULL"/>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<uses-permission android:name="android.permissions.ACCESS_WIFI_STATE" />
<uses-permission android:name="android.permissions.CHANGE_WIFI_STATE" />
<activity
android:name=".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>
</application>
编辑:更改后的 list :
<?xml version="1.0" encoding="utf-8"?>
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.INTERACT_ACROSS_USERS_FULL"/>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<uses-permission android:name="android.permissions.ACCESS_WIFI_STATE" />
<uses-permission android:name="android.permissions.CHANGE_WIFI_STATE" />
<application
android:allowBackup="true"
android:icon="@mipmap/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme"
android:sharedUserId="android.uid.system">
<activity
android:name=".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>
</application>
编辑:MainActivity 更新:
public class MainActivity extends Activity {
LinearLayout rl;
Button btn;
private TextView ErrorHandler;
private HttpPost httppost;
StringBuffer buffer;
HttpResponse response;
HttpClient httpclient;
List<NameValuePair> nameValuePairs;
TextView et,pass;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
rl = (LinearLayout) findViewById(R.id.rl);
et = new TextView(this);
pass = new TextView(this);
et.setVisibility(View.VISIBLE);
pass.setVisibility(View.VISIBLE);
et.setText("aici");
pass.setText("aici");
btn = new Button(this);
btn.setVisibility(View.VISIBLE);
btn.setText("Apasa!");
btn.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
login();
/* try {
OpenHttpConnection("http://78.97.183.34/srv.php");
} catch (IOException e) {
e.printStackTrace();
}*/
}
});
rl.addView(et);
rl.addView(pass);
rl.addView(btn);
}
private int login() {
int resp = 1;
try {
httpclient = new DefaultHttpClient();
httppost = new HttpPost("http://xx.xx.xx.xx/srv.php");
nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("username", et
.getText().toString().trim()));
nameValuePairs.add(new BasicNameValuePair("password", pass
.getText().toString().trim()));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
response = httpclient.execute(httppost);
ResponseHandler<String> responseHandler = new BasicResponseHandler();
final String response = httpclient.execute(httppost,
responseHandler);
if (response.equalsIgnoreCase("User Found")) {
et.setText("MERGE");
pass.setText("MERGE");
} else if (response.equalsIgnoreCase("No Such User Found")) {
resp = 0;
}
} catch (Exception e) {
System.out.println("Exception : " + e.getMessage());
}
return resp;
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.menu_main, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
//noinspection SimplifiableIfStatement
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
最佳答案
兄弟你的manifest文件好像有问题 将您的许可放在应用程序标签之上,如果问题仍然存在,请告诉我
[编辑]
<uses-permission android:name="android.permissions.ACCESS_WIFI_STATE" />
<uses-permission android:name="android.permissions.CHANGE_WIFI_STATE" />
将权限更改为权限并运行您的代码一次
<uses-permission android:name="android.permission.ACCESS_WIFI_STATE" />
<uses-permission android:name="android.permission.CHANGE_WIFI_STATE" />
关于Android令人沮丧的网络错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30364076/
大约一年前,我决定确保每个包含非唯一文本的Flash通知都将从模块中的方法中获取文本。我这样做的最初原因是为了避免一遍又一遍地输入相同的字符串。如果我想更改措辞,我可以在一个地方轻松完成,而且一遍又一遍地重复同一件事而出现拼写错误的可能性也会降低。我最终得到的是这样的:moduleMessagesdefformat_error_messages(errors)errors.map{|attribute,message|"Error:#{attribute.to_s.titleize}#{message}."}enddeferror_message_could_not_find(obje
我遵循MichaelHartl的“RubyonRails教程:学习Web开发”,并创建了检查用户名和电子邮件长度有效性的测试(名称最多50个字符,电子邮件最多255个字符)。test/helpers/application_helper_test.rb的内容是:require'test_helper'classApplicationHelperTest在运行bundleexecraketest时,所有测试都通过了,但我看到以下消息在最后被标记为错误:ERROR["test_full_title_helper",ApplicationHelperTest,1.820016791]test
我是rails的新手,想在form字段上应用验证。myviewsnew.html.erb.....模拟.rbclassSimulation{:in=>1..25,:message=>'Therowmustbebetween1and25'}end模拟Controller.rbclassSimulationsController我想检查模型类中row字段的整数范围,如果不在范围内则返回错误信息。我可以检查上面代码的范围,但无法返回错误消息提前致谢 最佳答案 关键是您使用的是模型表单,一种显示ActiveRecord模型实例属性的表单。c
我正在尝试编写一个将文件上传到AWS并公开该文件的Ruby脚本。我做了以下事情:s3=Aws::S3::Resource.new(credentials:Aws::Credentials.new(KEY,SECRET),region:'us-west-2')obj=s3.bucket('stg-db').object('key')obj.upload_file(filename)这似乎工作正常,除了该文件不是公开可用的,而且我无法获得它的公共(public)URL。但是当我登录到S3时,我可以正常查看我的文件。为了使其公开可用,我将最后一行更改为obj.upload_file(file
我克隆了一个rails仓库,我现在正尝试捆绑安装背景:OSXElCapitanruby2.2.3p173(2015-08-18修订版51636)[x86_64-darwin15]rails-v在您的Gemfile中列出的或native可用的任何gem源中找不到gem'pg(>=0)ruby'。运行bundleinstall以安装缺少的gem。bundleinstallFetchinggemmetadatafromhttps://rubygems.org/............Fetchingversionmetadatafromhttps://rubygems.org/...Fe
在Cooper的书BeginningRuby中,第166页有一个我无法重现的示例。classSongincludeComparableattr_accessor:lengthdef(other)@lengthother.lengthenddefinitialize(song_name,length)@song_name=song_name@length=lengthendenda=Song.new('Rockaroundtheclock',143)b=Song.new('BohemianRhapsody',544)c=Song.new('MinuteWaltz',60)a.betwee
我是Google云的新手,我正在尝试对其进行首次部署。我的第一个部署是RubyonRails项目。我基本上是在关注thisguideinthegoogleclouddocumentation.唯一的区别是我使用的是我自己的项目,而不是他们提供的“helloworld”项目。这是我的app.yaml文件runtime:customvm:trueentrypoint:bundleexecrackup-p8080-Eproductionconfig.ruresources:cpu:0.5memory_gb:1.3disk_size_gb:10当我转到我的项目目录并运行gcloudprevie
我有两个Rails模型,即Invoice和Invoice_details。一个Invoice_details属于Invoice,一个Invoice有多个Invoice_details。我无法使用accepts_nested_attributes_forinInvoice通过Invoice模型保存Invoice_details。我收到以下错误:(0.2ms)BEGIN(0.2ms)ROLLBACKCompleted422UnprocessableEntityin25ms(ActiveRecord:4.0ms)ActiveRecord::RecordInvalid(Validationfa
我想在Ruby中创建一个用于开发目的的极其简单的Web服务器(不,不想使用现成的解决方案)。代码如下:#!/usr/bin/rubyrequire'socket'server=TCPServer.new('127.0.0.1',8080)whileconnection=server.acceptheaders=[]length=0whileline=connection.getsheaders想法是从命令行运行这个脚本,提供另一个脚本,它将在其标准输入上获取请求,并在其标准输出上返回完整的响应。到目前为止一切顺利,但事实证明这真的很脆弱,因为它在第二个请求上中断并出现错误:/usr/b
这个问题在这里已经有了答案:Arraysmisbehaving(1个回答)关闭6年前。是否应该这样,即我误解了,还是错误?a=Array.new(3,Array.new(3))a[1].fill('g')=>[["g","g","g"],["g","g","g"],["g","g","g"]]它不应该导致:=>[[nil,nil,nil],["g","g","g"],[nil,nil,nil]]