java - Android Studio 不会发布到 localhost,也不会出错。
问题描述
我有一个基本的注册脚本。我花了很长时间才达到这一点,但我遇到了一个“错误”。该应用程序应该向 localhost phpmyadmin 发送 3 个登录注册字符串。该应用程序运行良好,直到它会生成错误弹出消息,我对其进行了编程。但它并没有告诉我为什么注册失败。
这是创建用户脚本。它应该从文本框中获取变量并将它们发送到 RegisterRequest 脚本。它等待响应,然后将用户踢回主菜单或创建重试消息。它会创建一条重试消息,但不会记录错误。
public class CreateUser extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_create_user);
this.setTitle("Create User");
final EditText username1 = findViewById(R.id.Createusername);
final EditText password1 = findViewById(R.id.CreatePassword);
final Switch isAdmin = findViewById(R.id.isadmin);
final Button createuser = findViewById(R.id.createuserbtn);
if (getIntent().hasExtra("com.example.northlandcaps.crisis_response")){
isAdmin.setVisibility(View.GONE);
}
createuser.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
final String username = username1.getText().toString();
final String password = password1.getText().toString();
final String isadmin = isAdmin.getText().toString();
Response.Listener<String> responseListener = new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Log.d("Response Value: ", response);
if (response.equals("success")){
Intent intent = new Intent(CreateUser.this, MainActivity.class);
CreateUser.this.startActivity(intent);
}else{
AlertDialog.Builder builder = new AlertDialog.Builder(CreateUser.this);
builder.setMessage("Register Failed")
.setNegativeButton("Retry",null)
.create()
.show();
}
}
};
RegisterRequest registerRequest = new RegisterRequest(username,password,isadmin,responseListener);
RequestQueue queue = Volley.newRequestQueue(CreateUser.this);
queue.add(registerRequest);
}
});
}
这是注册请求。它从 CreateUser 接收字符串并将它们发布到 php 脚本。我没有从这个脚本中得到任何错误,但可能有一些
public class RegisterRequest extends StringRequest {
private static final String REGISTER_REQUEST_URL = "http://192.168.*.*:80/phptesting/Register.php";
private Map<String, String> params;
public RegisterRequest(String username, String password,String isAdmin, Response.Listener<String> listener){
super(Method.POST, REGISTER_REQUEST_URL,listener,null);
params = new HashMap<>();
params.put("username",username);
params.put("password",password);
params.put("isAdmin",isAdmin+"");
}
public Map<String, String> getparams() {
return params;
}
}
这是注册 php 脚本。
<?php
$db_host = '192.168.*.*:3306';
$db_user = 'root';
$db_pass = '';
$db_name = 'test';
$con = mysqli_connect($db_host,'root',$db_pass,$db_name);
if($con){
echo "connection successful";
}
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$isAdmin ="no";
$username="fakename";
$password="fakepassword";
$isAdmin = isset($_POST["isAdmin"]);
$username = isset($_POST["username"]);
$password = isset($_POST["password"]);
echo $isAdmin;
echo $username;
echo $password;
$statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement,'ssi',$username,$password,$isAdmin);
mysqli_stmt_execute($statement);
if(!$statement) { printf("Prepare failed: %s\n", mysqli_error($con)); }
echo "success";
?>
感谢所有帮助,谢谢
(有人告诉我在 php 脚本中添加 Var_dump,然后向后工作。(感谢 riggs),所以我得到的结果是:connection successfularray(0) { } success
解决方案
改变
$isAdmin ="no";
$username="fakename";
$password="fakepassword";
$isAdmin = isset($_POST["isAdmin"]);
$username = isset($_POST["username"]);
$password = isset($_POST["password"]);
echo $isAdmin;
echo $username;
echo $password;
$statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement,'ssi',$username,$password,$isAdmin);
mysqli_stmt_execute($statement);
if(!$statement) { printf("Prepare failed: %s\n", mysqli_error($con)); }
echo "success";
至
if(isset($_POST["isAdmin"]) && isset($_POST["username"]) && isset($_POST["password"]))
{
$username = isset($_POST["username"]);
$password = isset($_POST["password"]);
$isAdmin = isset($_POST["isAdmin"]);
$statement = mysqli_prepare($con, "INSERT INTO cresidentials (username,password,isAdmin) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement,'ssi',$username,$password,$isAdmin);
mysqli_stmt_execute($statement);
if(!$statement)
{
printf("Prepare failed: %s\n", mysqli_error($con));
}
echo "success";
}
else
echo "values not set";
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