我的服务器中的 PHP 错误

Question

大家好，我在 Android Studio 中有这段 Java 代码：

private void loginUser(){
pd = ProgressDialog.show(LoginActivity.this, "", "Loading...");
StringRequest stringRequest = new StringRequest(Request.Method.POST, LOGIN_URL,
        new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                try {
                    JSONObject jsonResponse = new JSONObject(response);
                    boolean success = jsonResponse.getBoolean("success");
                    if (success) {
                        launchHomeScreen();
                        pd.dismiss();
                        Toast.makeText(LoginActivity.this,"Welcome back " + username,Toast.LENGTH_LONG).show();
                    }
                    else {
                        loginButton.setBackgroundColor(0x73000000);
                        Toast.makeText(LoginActivity.this,"Try again later!",Toast.LENGTH_LONG).show();
                        pd.dismiss();
                    }
                }
                catch (JSONException e) {
                    loginButton.setBackgroundColor(0x73000000);
                    e.printStackTrace();
                    Toast.makeText(LoginActivity.this,e.toString(),Toast.LENGTH_LONG).show();
                    pd.dismiss();
                }
            }
        },
        new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                loginButton.setBackgroundColor(0x73000000);
                Toast.makeText(LoginActivity.this,error.toString(),Toast.LENGTH_LONG).show();
                pd.dismiss();
            }
        }){
    @Override
    protected Map<String,String> getParams(){
        Map<String,String> params = new HashMap<>();
        params.put(KEY_USERNAME,username);
        params.put(KEY_PASSWORD,password);
        return params;
    }

};
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);

}

这在我的服务器上作为 php 代码：

<?php
$con = mysqli_connect("x", "x", "x", "x");

$username = $_POST["username"];
$password = $_POST["password"];

$statement = mysqli_prepare($con, "SELECT * FROM user WHERE username = ?");
mysqli_stmt_bind_param($statement, "s", $username);
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $colUsername, $colPassword);

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

while(mysqli_stmt_fetch($statement)){
    if (password_verify($password, $colPassword)) {
        $response["success"] = true;  
    }
}
header('Content-Type: application/json');
echo json_encode($response);

?>

这就是我从服务器得到的错误：

[29-Mar-2017 18:24:46 UTC] PHP Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't match number of fields in prepared statement in /home/u371648055/public_html/login.php on line 11

所以我不明白我的代码有什么问题。我知道 Javacode 是正确的....所以....这是 phpcode..但是什么？

我的 MySQL 数据库：

对于密码的问题 = 0 这是我发送到服务器的内容：

private void registerUser(){
    pd = ProgressDialog.show(RegisterActivity.this, "", "Loading...");
    StringRequest stringRequest = new StringRequest(Request.Method.POST, REGISTER_URL,
            new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    try {
                        JSONObject jsonResponse = new JSONObject(response);
                        boolean success = jsonResponse.getBoolean("success");
                        if (success) {
                            launchHomeScreen();
                            pd.dismiss();
                        }
                        else {
                            Toast.makeText(RegisterActivity.this,"Try again later!",Toast.LENGTH_LONG).show();
                            pd.dismiss();
                            Intent loginActivity = new Intent (RegisterActivity.this,LoginActivity.class);
                            RegisterActivity.this.startActivity(loginActivity);
                            finish();
                        }
                    }
                    catch (JSONException e) {
                        e.printStackTrace();
                        pd.dismiss();
                    }
                }
            },
            new Response.ErrorListener() {
                @Override
                public void onErrorResponse(VolleyError error) {
                    Toast.makeText(RegisterActivity.this,error.toString(),Toast.LENGTH_LONG).show();
                    pd.dismiss();
                }
            }){
        @Override
        protected Map<String,String> getParams(){
            Map<String,String> params = new HashMap<>();
            params.put(KEY_USERNAME,username);
            params.put(KEY_PASSWORD,password);
            params.put(KEY_MOBILENUMBER,mobileNumber);
            params.put(KEY_EMAIL, email);
            return params;
        }

    };
    RequestQueue requestQueue = Volley.newRequestQueue(this);
    requestQueue.add(stringRequest);

} 这就是注册 php 代码：

<?php
$connect = mysqli_connect("x", "x", "x", "x");

$username = $_POST["username"];
$password = $_POST["password"];
$mobilenumber = $_POST["mobilenumber"];
$email = $_POST["email"];

registerUser();

function registerUser() {
    global $connect, $username, $password, $mobilenumber, $email;
    $statement = mysqli_prepare($connect, "INSERT INTO user (username, password, mobilenumber, email) VALUES (?, ?, ?, ?)");
    mysqli_stmt_bind_param($statement, "siss", $username, $password, $mobilenumber, $email);
    mysqli_stmt_execute($statement);
    mysqli_stmt_close($statement);  

}

我如何获得 $password：

public class RegisterActivity extends AppCompatActivity {

   public static final String KEY_PASSWORD = "password";
   private String password;

  //When clicking the Regsiter-Button:
   pwd = (EditText) findViewById(R.id.editText2);
   password = pwd.getText().toString().trim();
   registerUser();
}

Answer 1

您的查询获得的列多于您传递给mysqli_stmt_bind_result() 函数的变量。更改您的选择查询以仅获取您需要的那些列（我猜是username 和password）：

$statement = mysqli_prepare($con, "SELECT username,password FROM user WHERE username = ?");

Answer 2

您需要将? 占位符的数量与替换变量的数量相匹配；只有这样才能准备执行语句：

SELECT * FROM user WHERE ? = ?;

Answer 3

嗯，这里的大部分答案都是正确的，我看到了你的评论。

如果您查看密码列，所有行都具有相同的值“零”。可能在调试的时候，你会发现你作为post数据发送的密码值与数据库中对应行的密码值不一致。

还要注意密码不应该是 int 类型，而 varchar 更适合用作密码。尝试在您的请求和数据库中的行中保持相同的密码。 此外，请在此处发布您从 Android 发送的值，如果您的问题仍然存在，请在此处发布 password_verify() 函数的值。