改造响应从PHP返回null

Question

由getValue引起的错误，它是检查查询的响应。邮递员给出了正确的答案，但在应用程序上启动时为空。

代码如下：

    private void loggedin() {
        FirebaseUser user = mAuth.getCurrentUser();
        String id = user.getUid();


        Retrofit retrofit = new Retrofit.Builder()
                .baseUrl(Base_URL)
                .addConverterFactory(GsonConverterFactory.create())
                .build();

        LoginApi api = retrofit.create(LoginApi.class);
        Call<Value> call = api.login(id);
        call.enqueue(new Callback<Value>() {
            @Override
            public void onResponse(Call<Value> call, Response<Value> response) {
                try {
                    String value = response.body().getValue();
                    if (value.equals("1")) {
                        Intent intent = new Intent(Login.this, MainActivity.class);
                        startActivity(intent);
                        finish();
                    } else {
                        Intent intent = new Intent(Login.this, Register.class);
                        startActivity(intent);
                        finish();
                    }
                }catch (Throwable throwable){
                    Toast.makeText(Login.this, "Error " + throwable, Toast.LENGTH_SHORT).show();
                }

            }
            @Override
            public void onFailure(Call<Value> call, Throwable t) {
                Toast.makeText(Login.this, "Error " + t, Toast.LENGTH_SHORT).show();
            }
        });
    }

这是LoginAPI：

    public interface LoginApi {
    @FormUrlEncoded
    @POST("/login.php")
    Call<Value> login(@Field("id") String id);}

这是值代码：

    @SerializedName("value")
    private String value;


    public String getValue() {
        return value;
    }

这是php代码：

<?php

if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    $response = array();
    $id = $_POST['id'];

    $dbhost = 'localhost';
    $dbuser = 'root';     // ini berlaku di xampp
    $dbpass = '';         // ini berlaku di xampp
    $dbname = 'ezparking';
    $connect = new mysqli($dbhost, $dbuser, $dbpass, $dbname);

    $sql = "SELECT * FROM customer WHERE id_cust ='$id'";
    $check = mysqli_fetch_array(mysqli_query($connect, $sql));
    if (isset($check)) {
        $response["value"] = 1;
        echo json_encode($response);
    } else {
        $response["value"] = 0;
        echo json_encode($response);
    }
    mysqli_close($connect);
}

这是邮递员的回复： Postman Response