Question

我想用电子邮件和密码登录数据库并获取信息。我不知道我的问题是什么，因为当点击登录按钮时，什么都没有。错误是：

*org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject*.

我的登录代码：

Button SignIn = (Button) findViewById(R.id.btn_signIn);
final EditText etEmail = (EditText) findViewById(R.id.etMail);
final EditText etPassword = (EditText) findViewById(R.id.etPassword);

SignIn.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View v) {
        email = etEmail.getText().toString();
        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) {
                        String name = jsonResponse.getString("name");
                        String surname = jsonResponse.getString("surname");

                        Intent intent = new Intent(LoginActivity.this, UsersMainActivity.class);
                        intent.putExtra("name", name);

                        LoginActivity.this.startActivity(intent);
                    } else {
                        AlertDialog.Builder builder = new AlertDialog.Builder(LoginActivity.this);
                        builder.setMessage("Login Failed Please Try again")
                                .setNegativeButton("Retry", null)
                                .create()
                                .show();
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }
            }

        };

        LoginRequest loginRequest = new LoginRequest(email, password, responseListener);
        RequestQueue queue = Volley.newRequestQueue(LoginActivity.this);
        queue.add(loginRequest);
    }
});

我的PHP代码：

<?php
$con = mysqli_connect("localhost","id1519330","****","id1519330");

    $email = $_POST["email"];
    $password = $_POST["password"];
  $statement = mysqli_prepare($con, "SELECT * FROM user WHERE email = ? and password = ? ");

    mysqli_stmt_bind_param($statement, "ss", $email,$password);
    mysqli_stmt_execute($statement);
    mysqli_stmt_store_result($statement);
    mysqli_stmt_bind_result($statement, $id, $name, $surname, $email, $password);

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

 while(mysqli_stmt_fetch($statement)){

            $response["success"] = true;  
            $response["name"] = $name;
            $response["surname"] = $surname;
            $response["email"] = $email;
            $response["password"] = $password;

    }
    echo json_encode($response);
?>

Answer 1

似乎您的回复是提供HTML代码（当您的请求未提供状态200时发生。通过使用调试器或打印响应来检查您的String响应，您将能够知道什么是错误的你的要求。

Answer 2

可能是你response没有正确的json样式来解析。请参阅this page 有关更多详细信息，请尝试调试以查看此response

的值

java.lang.String无法转换为JSONObject 2

2 个答案: