javascript - 我尝试存储在数据库中的任何内容都输入值 '0'

Question

我已经按照本教程制作了一个登录/注册应用程序: https://www.youtube.com/watch?v=QxffHgiJ64M&list=PLe60o7ed8E-TztoF2K3y4VdDgT6APZ0ka

一切正常，但后来我删除了“age”参数并插入了“lastname”和“email”参数。因此，每次我尝试注册用户时，“电子邮件”字段的值为“0”(即使我不使用 @ 或 .)

那是我的 Register.php 文件:

<?php
    require("Password.php");
    $connect = mysqli_connect("x", "x", "x", "x");
    
    $name = $_POST["name"];
    $lastname = $_POST["lastname"];
    $email = $_POST["email"];
    $password = $_POST["password"];
     function registerUser() {
        global $connect, $name, $lastname, $email, $password;
        $passwordHash = password_hash($password, PASSWORD_DEFAULT);
        $statement = mysqli_prepare($connect, "INSERT INTO user (name, lastname, email, password) VALUES (?, ?, ?, ?)");
        mysqli_stmt_bind_param($statement, "ssis", $name, $lastname, $email, $passwordHash);
        mysqli_stmt_execute($statement);
        mysqli_stmt_close($statement);     
    }
    function emailAvailable() {
        global $connect, $email;
        $statement = mysqli_prepare($connect, "SELECT * FROM user WHERE email= ?"); 
        mysqli_stmt_bind_param($statement, "s", $email);
        mysqli_stmt_execute($statement);
        mysqli_stmt_store_result($statement);
        $count = mysqli_stmt_num_rows($statement);
        mysqli_stmt_close($statement); 
        if ($count < 1){
            return true; 
        }else {
            return false; 
        }
    }
    $response = array();
    $response["success"] = false;  
    if (emailAvailable()){
        registerUser();
        $response["success"] = true;  
    }
    
    echo json_encode($response);
    ?>

这是我的 RegisterActivity.java 文件:

package com.example.gustavo.loginregister;

import android.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 android.content.Intent;

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 edtNome = (EditText) findViewById(R.id.edtNome);
        final EditText edtSobrenome = (EditText) findViewById(R.id.edtSobrenome);
        final EditText edtEmail = (EditText) findViewById(R.id.edtEmail);
        final EditText edtSenha = (EditText) findViewById(R.id.edtSenha);
        final Button btnRegistrar = (Button) findViewById(R.id.btnRegistrar);

        btnRegistrar.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                final String name = edtNome.getText().toString();
                final String lastname = edtSobrenome.getText().toString();
                final String email = edtEmail.getText().toString();
                final String password = edtSenha.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);
                            } else{
                                AlertDialog.Builder builder = new AlertDialog.Builder(RegisterActivity.this);
                                builder.setMessage("Register Failed")
                                        .setNegativeButton("Retry",null)
                                        .create()
                                        .show();
                                }
                        } catch (JSONException e) {
                            e.printStackTrace();
                        }
                    }
                };

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

那个是我的 RegisterRequest.java 文件:

package com.example.gustavo.loginregister;

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

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

public class RegisterRequest extends StringRequest {
    private static final String REGISTER_REQUEST_URL="https://wavecheckapp.000webhostapp.com/Register.php";
    private Map<String, String> params;

    public RegisterRequest(String name, String lastname, String email, String password, Response.Listener<String> listener){
        super(Method.POST, REGISTER_REQUEST_URL, listener, null);
        params = new HashMap<String, String>();
        params.put("name", name);
        params.put("lastname", lastname);
        params.put("email", email);
        params.put("password", password);
    }

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

我做错了什么？我已经检查了很多次...没有 INT 参数，一切都是 STRING。数据库在 utf8_unicode_ci 上。名字、姓氏和密码都被完美地存储着。数据库中参数为VARCHAR类型。

Answer 1

一个简单的错误，您将 mysqli_stmt_bind_param 中的电子邮件地址字段与 i 而不是 s 混淆了。

改变，

mysqli_stmt_bind_param($statement, "ssis", $name, $lastname, $email, $passwordHash);

为了，

mysqli_stmt_bind_param($statement, "ssss", $name, $lastname, $email, $passwordHash);

我不确定 mysqli 如何将参数转换为您提供的类型，但是当您运行以下代码时，输​​出为 0:

<?php
echo (int) 'test@test.com'; ---> 0
?>