从android应用程序插入数据到PostgreSQL数据库使用PHP web服务问题


Inserting data from android application to PostgreSQL database using PHP web service issue

我对我遇到的这个问题感到很头疼。我试图让用户从他们的android应用程序上传一些数据到我开发的网站服务。

数据将使用JSON和Android上传到PHP web服务,然后将数据"插入"到我的PostgreSQL数据库中。

我不确定逻辑错误在我的整个应用程序中,因为应用程序在运行时不会产生错误,但当我检查我的PostgreSQL服务器空间的数据库记录时,没有提交数据。

请参阅下面的代码我正在使用,请尽量帮助确定我在哪里走错了。我在谷歌上找过教程,但它们都是基于从PHP web服务到android应用程序的读取数据,但我希望从android应用程序发送原始数据。

DataPost Activity

public void postData() throws JSONException{
       Toast.makeText(DataSummary.this, "Done! Check your profile online to see your record.", Toast.LENGTH_LONG).show();
       Thread trd = new Thread(new Runnable(){
           public void run(){
             //Create a new HttpClient and Post Header
               HttpClient httpclient = new DefaultHttpClient();
               HttpPost httppost = new HttpPost("http://users.aber.ac.uk/dwd/mfb/php/jsonscript.php");
               JSONObject json = new JSONObject();
               Bitmap bitmapOrg = BitmapFactory.decodeResource(getResources(), i);
                ByteArrayOutputStream bao = new ByteArrayOutputStream();
                bitmapOrg.compress(Bitmap.CompressFormat.JPEG, 90, bao);
                byte[] ba = bao.toByteArray();
                String ba1=Base64.encodeToString(ba, i);
               try {
                   //JSON data:
                   json.put("photo", ba1.toString());
                   json.put("name", name);
                   json.put("description", description);
                   json.put("latitude", latitude);
                   json.put("longitude", longitude);
                   json.put("project", project);
                   json.put("owner", username);
                   JSONArray postjson = new JSONArray();
                   postjson.put(json);
                   //Post the data
                   httppost.setHeader("json", json.toString());
                   httppost.getParams().setParameter("jsonpost", postjson);
                   //Execute HTTP Post Request
                   System.out.println(json);
                   HttpResponse response = httpclient.execute(httppost);
                   //for JSON
                   if(response != null)
                   {
                       InputStream is = response.getEntity().getContent();
                       BufferedReader reader = new BufferedReader(new InputStreamReader(is));
                       StringBuilder sb = new StringBuilder();
                       String line = null;
                       try{
                           while((line = reader.readLine()) != null){
                               sb.append(line + "'n");
                           }
                       } catch (IOException e){
                           e.printStackTrace();
                       } finally {
                           try {
                               is.close();
                           } catch(IOException e){
                               e.printStackTrace();
                           }
                       }
                   }
               } catch(ClientProtocolException e){
                   e.printStackTrace();
               } catch (IOException e){
                   e.printStackTrace();
               } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
           }
       });
       trd.start();
   }
PHP的Webservice

<?php
session_start();
$conn = pg_connect("database_string");
//VARIABLES TO BE WRITTEN TO THE DATABASE
$photo = $_REQUEST["photo"];
echo $photo;
$binary=base64_decode($photo);
header('Content-Type: bitmap; charset=utf-8');
$name = json_decode(stripslashes($_POST["name"]));
$safe_name = pg_escape_string($name);
$desc = json_decode(stripslashes($_POST["description"]));
$safe_desc = pg_escape_string($desc);
$latitude = json_decode(stripslashes($_POST["latitude"]));
$longitude = json_decode(stripslashes($_POST["longitude"]));
$project = json_decode(stripslashes($_POST["project"]));
$owner = json_decode(stripslashes($_POST["owner"]));
$id = pg_query("SELECT * FROM users WHERE email = $owner");
$id_assoc = pg_fetch_assoc($id);
$id_res = $id_assoc['u_id'];
//SQL STATEMENT HERE FOR INSERT
$res = pg_query("INSERT INTO records (photo, name, description, latitude, longitude, project, owner) VALUES ('$photo', '$safe_name', '$safe_desc', '$latitude', '$longitude', '$project', '$id_res'");
pg_close($conn);
?>

在我的书中,任何可以提供一些建议/教程/代码解决方案的人都将是英雄!

SELECT查询返回任何东西吗?我不是PHP专家,但在我看来,你发送的变量是错误的,所以不应该有:

$id = pg_query("SELECT * FROM users WHERE email = $owner");

,

$id = pg_query("SELECT * FROM users WHERE email ='".$owner."'");

与INSERT查询类似。其他想法:

    当你只想要一列时,不要做SELECT *,它会变慢。例如,在9.2中使用仅索引扫描,您可以直接从索引(email,id) 返回id。
  • 如果您只想使用用户的id,最好将其放在插入查询的子查询中INSERT INTO records ( ... ,owner) VALUES (... ,(SELECT id FROM users WHERE email='".$owner."')")您甚至可以在末尾添加RETURNING owner,以便从插入查询中获得所有者id,如果您需要它在其他地方。