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
- 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。
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。