gpt4 book ai didi

php - 为什么 MySQL 不存储来自 Android 应用程序的数据

转载 作者:行者123 更新时间:2023-11-29 09:46:01 25 4
gpt4 key购买 nike

基本上所有其他字段(例如用户名、密码、全名...)都可以存储到 MySQL 中,但“联系人”和“电子邮件”字段则不行。我在这里附上了一些代码。请帮助我..

我尝试了几种方法来调整字段的顺序,但仍然得到相同的结果。我什至重新启动了 xampp,但仍然无法让它工作。

public class RegisterActivity extends AppCompatActivity {
private static final String KEY_STATUS = "status";
private static final String KEY_MESSAGE = "message";
private static final String KEY_FULL_NAME = "full_name";
private static final String KEY_USERNAME = "username";
private static final String KEY_PASSWORD = "password";
private static final String KEY_CONTACT = "contact";
private static final String KEY_EMAIL = "email";
private static final String KEY_EMPTY = "";
private EditText etUsername;
private EditText etPassword;
private EditText etConfirmPassword;
private EditText etFullName;
private EditText etContact;
private EditText etEmail;
private String username;
private String password;
private String confirmPassword;
private String fullName;
private String contact;
private String email;
private ProgressDialog pDialog;
private String register_url = "http://10.55.1.143/buildingreg/register.php";
private SessionHandler session;

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
session = new SessionHandler(getApplicationContext());
setContentView(R.layout.activity_register);

etUsername = findViewById(R.id.etUsername);
etPassword = findViewById(R.id.etPassword);
etConfirmPassword = findViewById(R.id.etConfirmPassword);
etFullName = findViewById(R.id.etFullName);
etContact = findViewById(R.id.etContact);
etEmail = findViewById(R.id.etEmail);

TextView login = findViewById(R.id.btnRegisterLogin);
Button register = findViewById(R.id.btnRegister);

//Launch Login screen when Login Button is clicked
login.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
Intent i = new Intent(RegisterActivity.this, LoginActivity.class);
startActivity(i);
finish();
}
});

register.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
//Retrieve the data entered in the edit texts
username = etUsername.getText().toString().toLowerCase().trim();
password = etPassword.getText().toString().trim();
confirmPassword = etConfirmPassword.getText().toString().trim();
fullName = etFullName.getText().toString().trim();
contact = etContact.getText().toString().trim();
email = etEmail.getText().toString().trim();
if (validateInputs()) {
registerUser();
}

}
});

}

private void registerUser() {
displayLoader();
JSONObject request = new JSONObject();
try {
//Populate the request parameters
request.put(KEY_USERNAME, username);
request.put(KEY_PASSWORD, password);
request.put(KEY_FULL_NAME, fullName);
request.put(KEY_CONTACT, contact);
request.put(KEY_EMAIL, email);

} catch (JSONException e) {
e.printStackTrace();
}
JsonObjectRequest jsArrayRequest = new JsonObjectRequest
(Request.Method.POST, register_url, request, new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
pDialog.dismiss();
try {
//Check if user got registered successfully
if (response.getInt(KEY_STATUS) == 0) {
//Set the user session
session.loginUser(username,fullName);
loadMain();

}else if(response.getInt(KEY_STATUS) == 1){
//Display error message if username is already existsing
etUsername.setError("Username already taken!");
etUsername.requestFocus();

}else{
Toast.makeText(getApplicationContext(),
response.getString(KEY_MESSAGE), Toast.LENGTH_SHORT).show();

}
} catch (JSONException e) {
e.printStackTrace();
}
}
}, new Response.ErrorListener() {

@Override
public void onErrorResponse(VolleyError error) {
pDialog.dismiss();

//Display error message whenever an error occurs
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_SHORT).show();

}
});

// Access the RequestQueue through your singleton class.
MySingleton.getInstance(this).addToRequestQueue(jsArrayRequest);
}

未显示错误,但数据库未显示从用户端捕获的“联系人”和“电子邮件”。

下面是注册用户的php代码。

<?php
$response = array();
include 'db/db_connect.php';
include 'functions.php';

//Get the input request parameters
$inputJSON = file_get_contents('php://input');
$input = json_decode($inputJSON, TRUE); //convert JSON into array

//Check for Mandatory parameters
if(isset($input['username'])&&isset($input['password'])
&&isset($input['contact'])&&isset($input['email'])
&&isset($input['full_name'])){
$username = $input['username'];
$password = $input['password'];
$fullName = $input['full_name'];
$contact = $input['contact'];
$email = $input['email'];

//Check if user already exist
if(!userExists($username)){

//Get a unique Salt
$salt = getSalt();

//Generate a unique password Hash
$passwordHash = password_hash(concatPasswordWithSalt($password,$salt),PASSWORD_DEFAULT);

//Query to register new user
$insertQuery = "INSERT INTO member(full_name, contact, email, username, password_hash, salt) VALUES (?,?,?,?)";
if($stmt = $con->prepare($insertQuery)){
$stmt->bind_param("ssss",$fullName,$contact,$email,$username,$passwordHash,$salt);
$stmt->execute();
$response["status"] = 0;
$response["message"] = "User created";
$stmt->close();
}
}
else{
$response["status"] = 1;
$response["message"] = "User exists";
}
}
else{
$response["status"] = 2;
$response["message"] = "Missing mandatory parameters";
}
echo json_encode($response);
?>

最佳答案

请尝试

$insertQuery  = "INSERT INTO member(full_name, contact, email, username, password_hash, salt) VALUES (?,?,?,?,?,?)";

而不是

$insertQuery  = "INSERT INTO member(full_name, contact, email, username, password_hash, salt) VALUES (?,?,?,?)";

您有 6 个参数,但只有 4 个参数。

关于php - 为什么 MySQL 不存储来自 Android 应用程序的数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55646169/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com