我在数据库表中插入数据时遇到问题,并将信息从android studio传递到数据库

时间:2017-02-26 00:47:59

标签: java php android mysql

下面是我的用户注册注册类,在用户点击注册的类中,它从字段中获取信息并将其传递给注册请求(下面发布的注册请求代码),我没有添加任何内容在android studio中执行我的代码后的表。

package com.example.user.emaillogin;

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 Register extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_register);
    final EditText etemail =(EditText) findViewById(R.id.etemail);
    final EditText etpassword =(EditText) findViewById(R.id.etpassword);
    final EditText etname =(EditText) findViewById(R.id.etname);
    final Button buttonregister =(Button) findViewById(R.id.buttonregister);

    buttonregister.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            final String name = etname.getText().toString();
            final String email = etemail.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(Register.this,MainActivity.class);
                            Register.this.startActivity(intent);
                        } else{
                            AlertDialog.Builder builder = new   AlertDialog.Builder(Register.this);
                            builder.setMessage("Register failed")
                                    .setNegativeButton("Retry", null)
                                    .create()
                                    .show();
                        }
                    } catch (JSONException e) {
                        e.printStackTrace();
                    }
                }
            };
            Registerrequest registerRequest = new Registerrequest(name,   email, password, ResponseListener);
            RequestQueue queue = Volley.newRequestQueue(Register.this);
            queue.add(registerRequest);
        }
    });
}
}

这个Register Request类允许我向服务器上的register.php(下面发布的代码)文件发出请求,并以字符串形式获取响应,并将php文件上传到我的服务器

package com.example.user.emaillogin;

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

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

/**
 * Created by User on 07/02/2017.
 */

public class Registerrequest extends StringRequest {

private static final String REGISTER_REQUEST_URL =    "http://picandlaugh.co.uk/register.php";
private Map<String, String> params;

public Registerrequest(String name, String email, String password,   Response.Listener<String> listener){

    super(Method.POST, REGISTER_REQUEST_URL, listener, null);
    params = new HashMap<>();
    params.put("name",name);
    params.put("email",email);
    params.put("password",password);
}


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


}

下面是我的register.php代码,我确实在实际代码中包含了实际的主机,用户名,密码等,但出于安全目的将其删除,

<?php
$con = mysqli_connect("HOST", "USERNAME",  "PASSWORD", "DB");

$name = $_POST["name"];
$email = $_POST["mail"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, " INSERT INTO WINKEL (name, mail, password) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement, "sss", $name, $email, $password);
mysqli_stmt_execute($statement);

$response = array();
$response["success"] = true;  
echo json_encode($response);
?>

0 个答案:

没有答案