我正在做一些分配,并且需要在本地主机服务器上启用SQLite DB数据同步到MySQL数据库。点击按钮时,来自SQLite的数据需要被“收集”,转换为JSON并发送到本地MySQL数据库并插入到那里。我做了一个Activity来处理这个工作,根据学院的例子做了一些PHP,并且让wamp服务器运行在我创建需要存储数据的数据库的地方。 在我的本地主机数据库上被命名为employes_db,并且内部的表被命名为employes。 下面是Android Studio中的代码,我提出:Android - 在本地主机服务器上发布从Android SQLite数据库到MySQL数据库的数据
package com.EmDatabase;
import android.app.Activity;
import android.content.Intent;
import android.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Toast;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class DataSyncManager extends Activity {
Database db;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.sync_options);
db = new Database(this);
db.getWritableDatabase();
}
public void syncToServer(View v) throws JSONException {
List<Employe> employes = db.selectAll();
JSONObject objEmployes = new JSONObject();
JSONArray employesData = new JSONArray();
for (Employe employe : employes) {
int id = employe.getId();
String name = employe.getName();
String surname = employe.getSurname();
int age = employe.getAge();
String company = employe.getCompany();
String wtype = employe.getWorktype();
JSONObject empData = new JSONObject();
empData.put("id", id);
empData.put("name", name);
empData.put("surname", surname);
empData.put("age", age);
empData.put("company", company);
empData.put("worktype", wtype);
employesData.put(empData);
}
objEmployes.put("all_employes", employesData);
String result = objEmployes.toString();
System.out.println(result);
UploadJsonStringTask task = new UploadJsonStringTask();
task.execute(
new String[] { "http://10.0.2.2/employes_db/register.php", result}
);
}
public void syncFromServer(View v) {
}
private class UploadJsonStringTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
String response = "";
Map<String,String> queries = new HashMap<String, String>(1);
queries.put("all_employes", params[1]);
try {
response += postHttpContent(params[0],queries);
} catch (IOException e) {
Log.e("error", e.toString());
}
return response;
}
@Override
protected void onPostExecute(String result) {
Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG).show();
}
public String postHttpContent(String urlString, Map<String, String> queries) throws IOException {
String response = "";
URL url = new URL(urlString);
HttpURLConnection httpConnection = (HttpURLConnection) url.openConnection();
httpConnection.setDoInput(true);
httpConnection.setDoOutput(true);
httpConnection.setUseCaches(false);
httpConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
String postData = "";
for (String key : queries.keySet()) {
postData += "&" + key + "=" + queries.get(key);
}
postData = postData.substring(1);
DataOutputStream postOut = new DataOutputStream(httpConnection.getOutputStream());
postOut.writeBytes(postData);
postOut.flush();
postOut.close();
int responseCode = httpConnection.getResponseCode();
if (responseCode == HttpURLConnection.HTTP_OK) {
String line;
BufferedReader br = new BufferedReader(new InputStreamReader(httpConnection.getInputStream()));
while ((line = br.readLine()) != null) {
response += line;
}
} else {
response = "";
throw new IOException();
}
return response + " *** Uploaded!";
}
}
public void goBack(View v) {
Intent in= new Intent(DataSyncManager.this,MainActivity.class);
startActivity(in);
}
}
这里是PHP文件,我做了,并插入到wampserver/WWW/employes_db(register.php):
<?php
$id = 0;
$name = "";
$surname = "";
$age = 0;
$company = "";
$worktype = "";
$conn = new mysqli("localhost","root","","employes_db");
$conn->query("insert into employes values (null,'".$_POST['id']."','".$_POST['name']."','".$_POST['surname']."','".$_POST['age']."','".$_POST['company']."','".$_POST['worktype']."')");
if(!$conn->error) echo "{status:0}"; else echo "{status:-1}";
?>
当我启动应用程序时,在SQLite数据库中插入一行,比命中同步按钮,并打开“localhost/employes_db/register.php”我得到“错误” - >注意:未定义的索引:id在C:\ wamp64 \ www \ employes_db \ register.php第9行< - 同样的错误对于休息列(姓名,年龄,公司,wtype)。 有人可以帮忙,我的错误在哪里?
我需要使用POST insted的因为我从我的数据库传输数据,所以我不会发送一个,两个或三个数据但更多,并且它依赖于我在点击同步按钮之前插入数据库的数据量。 你可以解释一下如何设置file_get_contents ...在我的情况下,以及在哪里,请注意我对PHP不太好,我只使用我在学院中找到的示例编写了我的任务代码。 – Mystiq
@Mystiq我添加了示例代码,多数民众赞成我如何解决你的问题,希望它可以帮助你,我正确理解你 – Tyrmos
我用你的PHP脚本,但保持我的代码从应用程序,因为当我把url.openConnection();它变红了,就像没有办法......我进一步了,但现在我又得到了一个“警告”......这里是: 注意:未定义变量:localhost在C:\ wamp64 \ www \ employes_db \ register .php on line 3 同样适用于root用户,也适用于密码...以及以下警告: 警告:mysqli_connect():(HY000/1045):拒绝用户'@'localhost'的访问(使用密码: NO)在第3行的C:\ wamp64 \ www \ employes_db \ register.php中 您对此有何想法? – Mystiq