使用parse-server-example和parse-dashboard连接新的Android项目(本地) [英] Connecting a new Android project with parse-server-example and parse-dashboard (local)
问题描述
我正在尝试将一个新的android项目与parse-server-example示例(本地)连接起来，并使用parse-dashboard显示它.我已经成功地将parse-server-example与parse-dashboard链接了...(通过更改parse-server-example/index.js和parse-dashboard/parse-dashboard-config.json中的appId和masterKey).我已经安装了mongodb和所有内容(mongodb，parse-server-example和parse-dashboard运行正常).
I am trying to connect a new android project with the parse-server-example (locally) and showing it with the parse-dashboard. I have successfully linked parse-server-example with parse-dashboard...(by changing appId and masterKey in parse-server-example/index.js and in parse-dashboard/parse-dashboard-config.json). I have installed mongodb and everything (mongodb,parse-server-example and parse-dashboard is working perfectly).
但是现在当我尝试在android项目中初始化解析服务器时，它不会在本地解析中创建任何内容(类).(在parse.com上运行良好).谢谢.我的密码
But now when i am trying to initialize parse server in android project, it does not create anything(class) in parse locally.(working fine with parse.com). Thanks. MY code
清单文件
<?xml version="1.0" encoding="utf-8"?>
<!--
~ Copyright (c) 2015-present, Parse, LLC.
~ All rights reserved.
~
~ This source code is licensed under the BSD-style license found in the
~ LICENSE file in the root directory of this source tree. An additional grant
~ of patent rights can be found in the PATENTS file in the same directory.
-->
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.parse.starter" >
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<application
android:name=".StarterApplication"
android:allowBackup="true"
android:icon="@mipmap/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme" >
<meta-data
android:name="com.parse.APPLICATION_ID"
android:value="@string/parse_app_id" />
<meta-data
android:name="com.parse.CLIENT_KEY"
android:value="@string/parse_client_key" />
<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>
</manifest>
Build.gradle文件
Build.gradle file
...
dependencies {
compile 'com.android.support:appcompat-v7:22.2.0'
compile 'com.parse.bolts:bolts-tasks:1.3.0'
compile 'com.parse:parse-android:1.13.0'
}
MainActivity.java
MainActivity.java
package com.parse.starter;
import android.os.Bundle;
import android.support.v7.app.ActionBarActivity;
import android.support.v7.app.AppCompatActivity;
import android.view.Menu;
import android.view.MenuItem;
import com.parse.Parse;
import com.parse.ParseAnalytics;
import com.parse.ParseObject;
public class MainActivity extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
ParseAnalytics.trackAppOpenedInBackground(getIntent());
}
@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);
}
}
StarterApplication.java
StarterApplication.java
package com.parse.starter;
import android.app.Application;
import com.parse.Parse;
import com.parse.ParseACL;
import com.parse.ParseObject;
import com.parse.ParseUser;
public class StarterApplication extends Application {
@Override
public void onCreate() {
super.onCreate();
// Enable Local Datastore.
Parse.enableLocalDatastore(this);
// Add your initialization code here
//Parse.initialize(this);
Parse.initialize(new Parse.Configuration.Builder(getApplicationContext())
.applicationId("@string/parse_app_id")
.clientKey("@string/parse_client_key")
.server("http://localhost:1337/parse/") // '/' important after 'parse'
.build());
ParseObject testObject = new ParseObject("TestObject");
testObject.put("foo", "bar");
testObject.saveInBackground();
ParseUser.enableAutomaticUser();
ParseACL defaultACL = new ParseACL();
// Optionally enable public read access.
// defaultACL.setPublicReadAccess(true);
ParseACL.setDefaultACL(defaultACL, true);
}
}
我也尝试过将对象的创建置于主要活动中
I have also tried putting the creation of object in main activtiy
MainActivity.java
MainActivity.java
public class MainActivity extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
ParseAnalytics.trackAppOpenedInBackground(getIntent());
ParseObject testObject = new ParseObject("TestObject");
testObject.put("foo", "bar");
testObject.saveInBackground();
}
推荐答案
为什么在代码中有localhost? Localhost表示该请求正在本地设备上运行，您可以在服务器内部使用它，但不能在客户端设备上使用它.尝试在其中放置服务器/计算机的一些IP地址
Why do you have localhost in your code? Localhost means that the request is running on local device, you can use it inside server but not on a client device. Try to put there some IP address of the server/your computer
.server("http://localhost:1337/parse/")
这篇关于使用parse-server-example和parse-dashboard连接新的Android项目(本地)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
