我正在嘗試開發一個簡單的Android應用程序,用於將用戶信息存儲到服務器上。所以我開發了我的應用程序的前端,並且在我的服務器上創建了一個MySql數據庫。我已經編寫了更新數據庫所需的PHP腳本,並且我在Android代碼中編寫了一個解析器來解析併發送數據到服務器。無法從Android代碼向PHP發送參數
問題是,我無法從我的Android代碼發送參數到我的PHP代碼。我已經閱讀了許多關於SO和許多其他論壇的問題和解答,但似乎沒有任何效果。請幫忙。
我創建了一個ASync任務,執行將數據發送到服務器的任務。以下是Async Task的代碼。
class CreateNewUser extends AsyncTask<String, String, String> {
@Override
protected void onPreExecute() {
super.onPreExecute();
}
// Creating user
protected String doInBackground(String... args) {
String name = user.getName();
String firstname = user.getFirstName();
String lastname = user.getLastName();
String number = user.getNumber();
String email = user.getEmail();
String status = user.getStatus();
String dob = user.getDob();
// Building Parameters
List<NameValuePair> params = new ArrayList<>();
params.add(new BasicNameValuePair("name", name));
params.add(new BasicNameValuePair("firstname", firstname));
params.add(new BasicNameValuePair("number", number));
params.add(new BasicNameValuePair("lastname", lastname));
params.add(new BasicNameValuePair("email", email));
params.add(new BasicNameValuePair("status", status));
params.add(new BasicNameValuePair("dob", dob));
JSONObject json = jsonParser.makeHttpRequest(url_create_product,
"POST", params);
try {
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
} else {
}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
protected void onPostExecute(String file_url) {
}
}
在上面的代碼中的「url_create_product」是URL到我的服務器的PHP創建用戶文件,這也是我以後會包括。
接下來是我的JSONParser文件。
package com.osahub.rachit.osachatting.server;
import android.util.Log;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.params.HttpConnectionParams;
import org.apache.http.params.HttpParams;
import org.json.JSONException;
import org.json.JSONObject;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;
/**
* Created by Rachit on 13-06-2015.
*/
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// Making HTTP request
try {
// check for request method
if (method == "POST") {
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
/*HttpParams httpParams = httpClient.getParams();
HttpConnectionParams.setConnectionTimeout(httpParams, 10000);
HttpConnectionParams.setSoTimeout(httpParams, 10000);*/
HttpPost httpPost = new HttpPost(url);
UrlEncodedFormEntity urlEncoded = new UrlEncodedFormEntity(params, "UTF-8");
httpPost.setEntity(urlEncoded);
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity entity = httpResponse.getEntity();
is = entity.getContent();
} else if (method == "GET") {
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
is = httpResponse.getEntity().getContent();
}
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json.substring(json.indexOf("{"), json.lastIndexOf("}") + 1));
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
最後,我的create_user.php文件在下面。
create_user.php
<?php
$response = array();
// check for required fields
if (isset($_POST['number'])) {
$name = $_POST['name'];
$first_name = $_POST['first_name'];
$number = $_POST['number'];
$last_name = $_POST['last_name'];
$email = $_POST['email'];
$status = $_POST['status'];
$dob = $_POST['dob'];
// include db connect class
require_once __DIR__ . '/user_info_connect.php';
// connecting to db
$db = new USER_INFO_CONNECT();
// mysql inserting a new row
$result = mysql_query("INSERT INTO user_info(name, first_name, last_name, number, email, status, dob) VALUES('$name', '$first_name', '$last_name', '$number', '$email', '$status', '$dob')");
// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["success"] = 1;
$response["message"] = "User successfully created.";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["success"] = 0;
$response["message"] = "Oops! An error occurred.";
// echoing JSON response
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
下面是MySQL數據庫中數據庫結構的快照。
此代碼的問題是,當我調試它時,查詢從未成功運行,它總是得到以下輸出「缺少必需的字段(s)」。這意味着數字字段丟失了,但爲了實驗的緣故,我甚至對用戶對象中的數字進行了硬編碼。我真的無法理解爲什麼這些參數沒有被代碼拾起。請幫忙。
我試着直接從瀏覽器執行這個查詢來測試我的php文件。 http://115.118.217.53:8068/osachat_connect/create_user.php?name=Rayzone&firstname=Ray&number=97179&lastname=zone&[email protected]&status=hoqdy&dob=3/7/89
即使在這裏我得到的迴應
create_user.php {"success":0,"message":"Required field(s) is missing"}
[您的腳本存在SQL注入風險。](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) –
如果可以,您應該[停止使用'mysql_ *'函數](http:// stackoverflow。COM /問題/ 12859942 /爲什麼 - 不應該,我使用MySQL的函數功能於PHP)。他們不再被維護,並[正式棄用](https://wiki.php.net/rfc/mysql_deprecation)。瞭解[準備](http://en.wikipedia.org/wiki/Prepared_statement)[聲明](http://php.net/manual/en/pdo.prepared-statements.php),並考慮使用PDO ,[這真的不難](http://jayblanchard.net/demystifying_php_pdo.html)。 –
我瞭解風險。這僅僅是爲了教育目的。我試圖找出爲什麼我的代碼沒有按照應有的方式工作(考慮到所有風險) – Rachit