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php - 从 Android 应用程序在 PHP 文件上发布 JSON 文件,并将 JSON 解码为 PHP 上的局部变量

转载 作者:行者123 更新时间:2023-11-29 02:52:46 25 4
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我已经对此进行了一段时间的研究,但一无所获。

我想做的是创建一个注册/登录 Activity ,它将所有访问详细信息存储在远程 SQL 数据库中。

我的代码大纲是创建“Registrar”对象,将其转换为 JSON 对象,并将该 JSON 对象转换为字符串,然后通过 httpclient 将该字符串作为帖子发送到 PHP 页面(位于在我的 XAMPP 上),请注意我使用的是 Android Studio 模拟器。

我的问题:

我不知道 JSON 文件是否被 PHP 服务器接收到。这是我的代码:

提交功能:

public void goSubmit(View view) throws IOException {

EditText nameEdit = (EditText) findViewById(R.id.nameEdit);
EditText idEdit = (EditText) findViewById(R.id.idEdit);
String name = nameEdit.getText().toString();
String ID = idEdit.getText().toString();

//Creating Student (Registrar) Object
Student registrar = new Student();
registrar.setMajor(majorEdit);
registrar.setName(name);
registrar.setId(ID);

//Creating JSON String


String registrarJSON = null;

try {
registrarJSON = ObjInJSON(registrar);
Toast.makeText(this, registrarJSON, Toast.LENGTH_LONG).show();
} catch (JSONException e) {
e.printStackTrace();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}

//Posting JSON String on Remote PHP

String PHPresponse = sendToRegistrationPHP(registrarJSON);
Toast.makeText(this, PHPresponse, Toast.LENGTH_LONG).show();
//Receive PIN from PHP as JSON String


//Parsing JSON string to integer (pin)


//Set PIN in registrar.getpin()


//Passing the object to setPassword Activity condition registrar.pin =! null


}

学生类(class):

 public class Student {

String Id = "NULL" ;
String Major = "NULL";
String Name = "NULL";
String Password = "NULL";
String Pin = "NULL";


public String getPin() {
return Pin;
}

public void setPin(String pin) {
Pin = pin;
}

public String getPassword() {
return Password;
}

public void setPassword(String password) {
Password = password;
}

public String getId() {
return Id;
}

public String setId(String id) {
Id = id;
return id;
}

public String getName() {
return Name;
}

public void setName(String name) {
Name = name;
}

public String getMajor() {
return Major;
}

public void setMajor(String major) {
Major = major;
}



}

创建字符串格式的 JSON 对象:

    protected String ObjInJSON(Student studentC) throws JSONException, UnsupportedEncodingException {
String ID = studentC.getId();
String Pin = studentC.getPin();
String Major = studentC.getMajor();
String Password = studentC.getPassword();
String Name = studentC.getName();


JSONObject json_obj = new JSONObject();

json_obj.put("id", ID);
json_obj.put("password", Password);
json_obj.put("pin", Pin);
json_obj.put("major", Major);
json_obj.put("name", Name);

return json_obj.toString();
}

发送到 PHP 服务器:

   public static String sendToRegistrationPHP(String jarr) throws IOException {
StringBuffer response = null;
try {

String myurl = "10.0.2.2:8070/StudentaccistancePHP/MySqlTEST.php";
URL url = new URL(myurl);

HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setReadTimeout(10000);
conn.setConnectTimeout(15000 /* milliseconds */);
conn.setRequestProperty("Content-Type", "application/json");
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setRequestMethod("POST");
OutputStream out = new BufferedOutputStream(conn.getOutputStream());
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(out, "UTF-8"));
writer.write(jarr);
writer.close();
out.close();

BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String inputLine;
response = new StringBuffer();

while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();

System.out.println("Response in universal: " + response.toString());
} catch (Exception exception) {
System.out.println("Exception: " + exception);
}

if (response != null) {
return response.toString();
}
else return "Not WORKING !";
}

PHP 服务器:

<?php


$json = file_get_contents('php://input');

$data = json_decode($json, true);
$ID = $data['id'];
$password = $data['password'];
$pin = "323232";
$major = $data['major'];
$name = $data['name'];


$servername = "localhost";
$username = "root";
$password = "";
$dbname = "studentassictance";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);

// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";

$sql = "INSERT INTO students (id, major, name, password, pin)
VALUES ('$ID', '$major', '$name', '$password', '$pin')";

if ($conn->query($sql) === TRUE) {
echo "New record created successfully <br>";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}




?>

但是,没有任何内容被插入到数据库中。

最佳答案

首先,您需要检查以 JSON 形式到达 PHP 端的内容。你可以

var_dump($json,$data);

json_encode() 调用后观察它是一个有效的 JSON。你可以验证它here

其次,向您展示SHOW CREATE TABLE students

第三,将所有内容重写为 PDO,因为它支持命名参数,如果需要的话,理论上您以后可以更轻松地迁移到另一个数据库引擎。所以它会是这样的:

<?php


define('DSN','mysql:dbname=test;host=localhost');
define('DB_USERNAME','testuser');
define('DB_PASSWORD','testpassword');

$connect = new PDO(DSN, DB_USERNAME, DB_PASSWORD);

$json = file_get_contents('php://input');
/*$json = '{
"id": "111",
"password": "sfsdfsdf",
"major": "Math",
"name": "Test User"
}';*/

$data = json_decode($json, true);

$ID = $data['id'];
$password = $data['password'];
$pin = "323232";
$major = $data['major'];
$name = $data['name'];

$sql = "INSERT INTO `students`(`id`,`major`, `name`, `password`, `pin`) VALUES(:id, :major, :name, :password, :pin)";

$result = $connect->prepare($sql);

//bind parameter(s) to variable(s)
$result->bindParam( ':id', $ID, PDO::PARAM_INT );
$result->bindParam( ':major', $major, PDO::PARAM_STR );
$result->bindParam( ':name', $name, PDO::PARAM_STR );
$result->bindParam( ':password', $password, PDO::PARAM_STR );
$result->bindParam( ':pin', $pin, PDO::PARAM_STR );

$status = $result->execute();

if ($status)
{
echo "New record created successfully <br>";

} else
{
echo "Error: <br>" .
var_dump($connect->errorInfo(),$status);
}

$connect = null;

关于php - 从 Android 应用程序在 PHP 文件上发布 JSON 文件,并将 JSON 解码为 PHP 上的局部变量,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33972814/

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