我试图用php +一个mysql数据库/ java for android创建一个简单的登录系统。 用于使用Json进行通信。
我在这里有PHP代码:(我想成为的一个简单例子）。
<?php
include("../dbfunctions.php");
class mobilelogin{
private $user;
private $pass;
public function getData(){
$get->user = $_POST['user'];
$get->pass = $_POST['pass'];
return $get;
}
public function getDbData($user){
$dbfunctions = new dbfunctions;
$getvalue = $this->getData();
return $dbfunctions->returnUser($user); // get pass by user
}
public function md5convert($user){
return md5($this->getDbData($user)); // convert this pass to md5
}
public function check(){ // check everthing here
$getvalue = $this->getData();
if(isset($getvalue->user) && isset($getvalue->pass)) // everthing filled in
{
return "valid";
}else{
return "invalid";
}
}
public function response(){ // show data here in json format
echo json_encode(array('result' => $this->check()));
}
}
header("Content-Type: text/json");
$mobilelogin = new mobilelogin;
echo $mobilelogin->response();
?>
就像你看到的那样，我只是尝试一个简单的json帖子示例，
我在json中的结果是：{ result:invalid }
如你所见，如果php代码从java接收post值，它应该显示有效，否则无效。
现在我有了这个Android代码，它发出了一个帖子请求：
public void login(View v) throws ClientProtocolException, URISyntaxException, IOException, JSONException
{
HttpPost httppost = new HttpPost(processUrl);
try {
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("user", uservalidated));
nameValuePairs.add(new BasicNameValuePair("pass", passvalidated));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
client.execute(httppost);
Toast.makeText(getBaseContext(),"Everthing well.",Toast.LENGTH_LONG).show();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
Log.e("testnow", response());
}
单击登录按钮时会发出post请求，php服务器应该接收它，
就像你看到Log（Log.e("testnow", response());）是通过httpget接收数据的函数，这个函数：
public String response() throws URISyntaxException, ClientProtocolException, IOException, JSONException{
HttpGet request = new HttpGet();
request.setURI(new URI(processUrl));
HttpResponse response;
response = client.execute(request);
BufferedReader in = new BufferedReader(new InputStreamReader(response
.getEntity().getContent()));
StringBuilder builder = new StringBuilder();
String line;
while((line=in.readLine())!=null)
{
builder.append(line);
}
String JSONdata = builder.toString();
JSONObject jObject = new JSONObject(JSONdata);
String JSONresponse = jObject.getString("result");
return JSONresponse;
}
这从json输出显示无效。但是当我在用户名和密码中勾选时，它应该显示有效，但它不会更新。
我希望你理解我的问题，欢迎所有的帮助。 如果这不是以这种方式建立登录系统的正确方法，也请分享。
答案 0 :(得分：2)
虽然我是Android应用程序开发人员，但我解决了很多问题。根据我的早期观点，我认为Android端没有问题，这是PHP端的问题。在您的服务器端，此代码不会更新代码数据。请验证您的代码。请看这个PHP方面的代码，这样你就可以创建以及更新代码了。
<?php
/*
* Following code will create a new profile row
* All profile details are read from HTTP Post Request
*/
// array for JSON response
$response = array();
// check for required fields
if (isset($_POST['name']) && isset($_POST['email']) && isset($_POST['mobile']) && isset($_POST['gender'])&& isset($_POST['location']) && isset($_POST['image']) && isset($_POST['fb_user']) && (trim($_POST['name']) != '') && (trim($_POST['email']) != '') && (trim($_POST['mobile']) != '') && (trim($_POST['gender']) != '') && (trim($_POST['location']) != '') && (trim($_POST['image']) != '') ) {
$name = $_POST['name'];
$email = $_POST['email'];
$mobile = $_POST['mobile'];
$gender = $_POST['gender'];
$country = $_POST['country'];
$location = $_POST['location'];
$image = $_POST['image'];
$fb_user = $_POST['fb_user'];
if($fb_user == '0'){
$pin = $_POST['pin'];
} else{
$pin = '0';
}
$created_date = date("Ymd");
$created_date2 = time();
$binary=base64_decode($image);
// binary, utf-8 bytes
// include db connect class
require_once 'db_connect.php';
// connecting to db
$db = new DB_CONNECT();
$chk_existing_user = "SELECT * FROM `user_profile` WHERE `email` ='".$email."'";
$res_existing_user = mysql_query($chk_existing_user);
$num_rows_existing_user = mysql_num_rows($res_existing_user); //checking for existing user
// mysql inserting a new row
header('Content-Type: image/jpg; charset=utf-8');
$username_arr = explode("@",$email);
$username = $username_arr[0].'_'.$created_date;
$file_name = 'user_images/'.$username.'.jpg';
$file = fopen($file_name, 'wb');
fwrite($file, $binary);
fclose($file);
if($num_rows_existing_user == 0){
$result = mysql_query("INSERT INTO `user_profile`(name, email, mobile, gender, country, location, image, pin, fb_user, created_date ) VALUES('$name', '$email' , '$mobile', '$gender' , '$country', '$location' , '$file_name', '$pin', '$fb_user', '$created_date')");
$inserted_edited = '1';
}else{
$result = mysql_query("UPDATE `user_profile` SET name='$name', mobile ='$mobile', gender ='$gender', country='$country', location ='$location', image = '$file_name', pin = '$pin', fb_user ='$fb_user', created_date ='$created_date2' WHERE email = '$email'");
$inserted_edited = '2';
}
// check if row inserted or not
if ($inserted_edited == '1') {
// successfully inserted into database
$response["success"] = 1;
$response["message"] = "profile successfully created!";
// echoing JSON response
echo json_encode($response);
} else if($inserted_edited == '2') {
// failed to insert row
$response["success"] = 2;
$response["message"] = "profile updated successfully!";
// echoing JSON response
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "required field(s) is missing!";
// echoing JSON response
echo json_encode($response);
}
?>
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享受!!!