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java - 使用 PHP 将数据从 Android 发送到 SQL 数据库

转载 作者:太空宇宙 更新时间:2023-11-04 11:55:50 27 4
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我正在按照本教程构建登录和注册应用程序:

https://www.youtube.com/watch?v=T7Z4GVFaT4A

我已经更改了 PHP 并通过地址字段传递一些参数来测试它,但是当我单击注册按钮时没有任何反应。OnResponse 函数中有些事情应该做一些事情,应该将一些东西添加到数据库中,但什么也没有发生。

我想知道如果我没有收到任何响应,是否可能会从应用程序发送任何内容。

此外,令我困扰的是,函数 OnResponse 从未被调用,但它应该被调用。

我使用 WAMP 创建服务器,而不是使用某些 ASP。下面是对此问题至关重要的 Java 和 PHP 文件。

RegisterRequest.java

package com.example.dominik.praca;

import com.android.volley.Response;
import com.android.volley.toolbox.StringRequest;

import java.util.HashMap;
import java.util.Map;

/**
* Created by Dominik on 28.12.2016.
*/

public class RegisterRequest extends StringRequest {
//private static final String REGISTER_REQUEST_URL = "http://kulturnik.ugu.pl/Register.php";
//private static final String REGISTER_REQUEST_URL = "http://kulturnik.byethost3.com/Register2.php";
private static final String REGISTER_REQUEST_URL = "http://127.0.0.1/kulturnik/Register.php";
private Map<String, String> params;
public RegisterRequest(String name, String username, String password, Response.Listener<String> listener){
super(Method.POST, REGISTER_REQUEST_URL, listener, null);
params = new HashMap<>();
params.put("name", name);
params.put("username", username);
params.put("password", password);
}

@Override
public Map<String, String> getParams() {
return params;
}
}

RegisterActivity.java

package com.example.dominik.praca;

import android.content.Intent;
import android.support.v7.app.AlertDialog;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;

import com.android.volley.RequestQueue;
import com.android.volley.Response;
import com.android.volley.toolbox.Volley;

import org.json.JSONException;
import org.json.JSONObject;

public class RegisterActivity extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_register);

final EditText etUsername = (EditText) findViewById(R.id.etUsername);
final EditText etPassword = (EditText) findViewById(R.id.etPassword);
final EditText etName = (EditText) findViewById(R.id.etName);
final Button bRegister = (Button) findViewById(R.id.bRegister);

bRegister.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
final String name = etName.getText().toString();
final String username = etUsername.getText().toString();
final String password = etPassword.getText().toString();

Response.Listener<String> responseListener = new Response.Listener<String>() {

@Override
public void onResponse(String response) {
try {
JSONObject jsonResponse = new JSONObject(response);
boolean success = jsonResponse.getBoolean("success");

if (success) {
Intent intent = new Intent(RegisterActivity.this, LoginActivity.class);
RegisterActivity.this.startActivity(intent);
AlertDialog.Builder builderSuccess = new AlertDialog.Builder(RegisterActivity.this);
builderSuccess.setMessage("Rejestracja zakończona powodzeniem")
.create()
.show();
}
else {
AlertDialog.Builder builder = new AlertDialog.Builder(RegisterActivity.this);
builder.setMessage("Błąd rejestracji")
.setNegativeButton("Ponów", null)
.create()
.show();
}
} catch (JSONException e) {
e.printStackTrace();
}

}
};

RegisterRequest registerRequest = new RegisterRequest(name, username, password, responseListener);
RequestQueue queue = Volley.newRequestQueue(RegisterActivity.this);
queue.add(registerRequest);
}
});
}
}

并注册.php

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

$con = mysqli_connect("localhost", "root", "", "kulturnik");

$name = $_POST["name"];
$username = $_POST["username"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO user (name, username, password) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement, "sss", $name, $username, $password);

mysqli_stmt_execute($statement);

$response = array();
$response["success"] = true;

print_r(json_encode($response));
?>

最佳答案

使用异步任务可以更快地向服务器发送和检索。只需创建该类的一个对象,然后使用执行函数传递参数并启动与服务器的异步发送和检索。

例如:- SendtoPhp stp = new SendtoPhp(); stp.execute(tname, 电话)

public class SendtoPhp extends AsyncTask<String, Void, String> {


@Override
protected String doInBackground(String... params) {
String tname = params[0]; //parameters you need to send to server
String phone = params[1];


String data = "";

int tmp;
try {

URL url = new URL("http://your.server.com/"); //url to php code
String urlParams = "tname=" + tname + "&tphone=" + phone;


HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setDoOutput(true);
OutputStream os = httpURLConnection.getOutputStream();

os.write(urlParams.getBytes());
os.flush();
os.close();

InputStream is = httpURLConnection.getInputStream();


while ((tmp = is.read()) != -1) {


data += (char) tmp;

//System.out.println(data);
}

is.close();
httpURLConnection.disconnect();

return data;

} catch (MalformedURLException e) {
e.printStackTrace();
return "Exception: " + e.getMessage();
} catch (IOException e) {
e.printStackTrace();
return "Exception: " + e.getMessage();
}
}

@Override
protected void onPostExecute(String s) {
try {
//You might receive something on server code execution using JSON object


} catch (JSONException e) {
e.printStackTrace();
}


}
}

关于java - 使用 PHP 将数据从 Android 发送到 SQL 数据库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41404949/

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