Spring boot application I can not get data from oracle database it returns []. In postman, it returns other requests e.g home method in controller class returns correctly. also, the table created by model class the problem is getting data from the table.
Here is the postman result:
I get this in console:
Model class
#Entity // This tells Hibernate to make a table out of this class
public class Userr {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private Integer id;
private String name;
private String email;
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
}
//Controller Class
#RestController
public class MainController {
#Autowired // This means to get the bean called userRepository
// Which is auto-generated by Spring, we will use it to handle the data
private UserRepository userRepository;
#PostMapping(path="/add") // Map ONLY POST Requests
public #ResponseBody String addNewUser (#RequestParam String name
, #RequestParam String email) {
// #ResponseBody means the returned String is the response, not a view name
// #RequestParam means it is a parameter from the GET or POST request
Userr n = new Userr();
n.setName(name);
n.setEmail(email);
userRepository.save(n);
return "Saved";
}
#GetMapping(path="/all")
public #ResponseBody Iterable<Userr> getAllUsers() {
// This returns a JSON or XML with the users
//
return userRepository.findAll();
}
#GetMapping(path="/al")
public List<Userr> printPersonInfo() {
List<Userr> list = new ArrayList<>();
userRepository.findAll().forEach(list::add);
return list;
}
#RequestMapping("/user")
public String home(){
return "PPPPPP";
}
}
//Repository Class
public interface UserRepository extends CrudRepository<Userr, Integer> {
}
Add #Repository annotation to your UserRepository. It will help with your issue.
Related
I am sending a JSON post request from postman in below format {"name":"ABC","id":1}
and trying to map it with my Employee class in Controller with #Requestbody
but still getting error as HTTP Status 415 – Unsupported Media Type with Headers ( Content-Type : application/json )
1) Controller :
#Controller
public class RegistrationController {
#Autowired
private EmployeeService employeeService;
#RequestMapping(path="/jsonreq", method=RequestMethod.POST, consumes=MediaType.APPLICATION_JSON_VALUE)
public String json(#RequestBody Employee e)
{
System.out.println("JSON_2::");
return "Testing";
}
}
2) Employee Class:
public class Employee {
private int id;
private String name;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
I want to be list out all my users with criteria of where id = formId. The code is working but just that it list out all the users instead of being filtered by formId. Please tell me where i did wrongly. Do tell me if you need any more info to solve this!
controller
*url = http://localhost:8080/User/Panda?Id=1
#RequestMapping(value = {"/{name}?Id={id}" }, method = RequestMethod.GET)
public String listClinicUser(ModelMap model, #PathVariable("id") Integer id) {
logger.info("Users List Page - Id = " + id);
List<User> user = service.findAllUsers(id);
model.addAttribute("users", user);
return "user/list";
}
Service
public List<User> findAllUsers(Integer id) {
return dao.findAllUsers(id);
}
DAO Class
public interface UserDao {
List<User> findAllUsers(Integer id);
}
*DAOImpl Class
#SuppressWarnings("unchecked")
public List<User> findAllUsers(Integer id) {
Criteria crit = createEntityCriteria();
crit.add(Restrictions.eq("formId",id));
crit.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);
List<User> users = (List<Usert>) crit.list();
return users;
}
*for createEntityCriteria() i created in another class call abstractDao and extends to it.
private final Class<T> persistentClass;
#SuppressWarnings("unchecked")
public AbstractDao(){
this.persistentClass =(Class<T>) ((ParameterizedType) this.getClass().getGenericSuperclass()).getActualTypeArguments()[1];
}
protected Criteria createEntityCriteria(){
return getSession().createCriteria(persistentClass);
}
Class Entity
#Entity
#Table(name="USER")
public class User implements Serializable{
#NotEmpty
#Column(name="formId", nullable=false)
private Integer formId;
#NotEmpty
#Column(name="FIRST_NAME", nullable=false)
private String firstName;
#NotEmpty
#Column(name="LAST_NAME", nullable=false)
private String lastName;
public Integer getFormId() {
return formId;
}
public void setFormId(Integer formId) {
this.formId= formId;
}
...
}
value = {"/{name}?Id={id}" }
This is wrong way to extract URL param. If you want to get URL param, you should pass it to your method using #RequestParam annotation:
#RequestMapping(value = {"/{name}" }, method = RequestMethod.GET)
public String listClinicUser(ModelMap model, #RequestParam("Id") Integer id) {
//...
}
Spring automatically pass value that you need. For example in case of ?Id=1 Spring will pass 1 to your controller
In your url /{name} is a path variable and is annotated with #PathVariable like in:
#RequestMapping(value = "/foo/bar/{name}", method = GET)
#ResponseBody
public String getBarByName(#PathVariable String name) { ... }
And ?Id=id is a request parameter and is annotated wiht #RequestParam so if we map to url like this one:
http://localhost:8080/api/foo/bar?id=100
we do it like this
#RequestMapping(value = "/foo/bar", method = GET)
#ResponseBody
public String getBarById(#RequestParam("id") Integer id) { ... }
So to combine them to map to your url:
#RequestMapping(value = {"/{name}" }, params = "id", method = RequestMethod.GET)
public String listClinicUser(ModelMap model, #PathVariable String name, #RequestParam("id" Integer id)) { ... }
How to handle exception handling in Spring Boot 1.5.4 without controller class? Currently, I have only entity & repository class as below.
Task.class: (entity)
#Entity
#Table(name = "task")
public class Task implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id")
private long id;
#Length(min = 1)
private String name;
public Task() {
}
public Task(String name) {
this.name = name;
}
public Task(Long id, String name) {
this.name = name;
}
public long getId() {
return id;
}
public String getName(){
return name;
}
}
Repository.class:
public interface TaskRepository extends PagingAndSortingRepository<Task, Long> {
}
POST method: return 200 ok
http://localhost:8080/tasks
{
"name" : "test"
}
But,
{
"name" : ""
}
returns 500 , instead of 400 error.
Pls let me know, if any way to handle this exception without a controller class.
You could use a global #ExceptionHandler with the #ControllerAdvice annotation. Basically, you define which Exception to handle with #ExceptionHandler within the class with #ControllerAdvice annotation, and then you implement what you want to do when that exception is thrown.
Like this:
#ControllerAdvice(basePackageClasses = RepositoryRestExceptionHandler.class)
public class GlobalExceptionHandler {
#ExceptionHandler({ValidationException.class, JsonParseException.class})
public ResponseEntity<Map<String, String>> yourExceptionHandler(Exception e) {
Map<String, String> response = new HashMap<String, String>();
response.put("message", "Bad Request");
return new ResponseEntity<Map<String, String>>(response, HttpStatus.BAD_REQUEST);
}
}
See also: http://www.ekiras.com/2016/02/how-to-do-exception-handling-in-springboot-rest-application.html
My problem is when I'm trying to update entity using GraphRepository 'save' method, all my fields, that was not changed, are being rewritten with null values.
I make PUT request with only id and name, because I want change User's name, but save method automatically changes password field to null value. I wanna update only dirty values. How can I achieve this? Maybe there is some kind of merge or update methods?
My entity class looks like this:
#NodeEntity
public class User {
#GraphId Long id;
String name;
String password;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#JsonIgnore
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
}
My repository class looks like this:
public interface UserRepository extends GraphRepository<User> {
public User findById(Long id);
}
#Transactional
public void update(Long id, String name) {
User user = repository.findById(id);
if (user != null) {
user.setName(name);
}
}
I am using spring data jpa with hibernate
This is my dao interface
#Repository
public interface IUserDAO extends JpaRepository<User, Integer>{
User findByUsername( final String username );
}
This is my User class
Entity
#Table(name="USER")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name="ID", nullable = false)
private int id;
#Column(name="USERNAME", nullable = false)
private String username;
#Column(name="NAME", nullable = false)
private String name;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
This is my UserImplClass
This is my UserImplClass{
#Autowired
private IUserDAO iUserDAO;
public String findUserByUserName(String username) {
User user =iUserDAO.findByUsername(username);
Convert user to json object from framework level automatically
// i can add my one implemenation of converting user to json here ,but i want to achieve it from framework so that my code is not scattered on every service level
return "jsonStringOfUserObject"
}
Is it possible with spring data jpa with hibernate so that i do not have to write code for converting java object to json string in every service level?
I am using spring ,therefore i want to achieve it from spring .
You have two options to do what you want:
1) If you plan on returning this Object as an HTTP Response, and you use Spring MVC with Controllers you can annotate your controller method as follows:
public #ResponseBody User getUser(){
return userImplClass.findUserByUserName("yourusername");
}
2) If you want the UserImplClass itself to return a JSON String (which I do't recommend, but I leave you the decision), you can use Jackson Object Mapper to do it for you (you can inject it if you declare it as a bean on your configuration xml, or create a new instance of it, I personally prefer injecting it with #Autowired)
public String findUserByUserName(String username) {
User user =iUserDAO.findByUsername(username);
ObjectMapper mapper = new ObjectMapper(); // no need to do this if you inject via #Autowired
return mapper.writeValueAsString(user);
}