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how to catch sqlException or signal message from mysql trigger
I want to catch message of a trigger from mysql database and print it via my thymeleaft template but I came across a problem with catching message of that trigger
part of pet repository
void save(Pet pet) throws SQLException;
this is part of a controller, method of catching doesnt work and I don't know how to do it properly
@PostMapping("/pets/new")
public String processCreationForm(Owner owner, @Valid Pet pet, BindingResult result, ModelMap model) {
if (StringUtils.hasLength(pet.getName()) && pet.isNew() && owner.getPet(pet.getName(), true) != null){
result.rejectValue("name", "duplicate", "already exists");
}
owner.addPet(pet);
if (result.hasErrors()) {
model.put("pet", pet);
return VIEWS_PETS_CREATE_OR_UPDATE_FORM;
} else {
try {
this.pets.save(pet);
}catch (SQLException e){
System.out.println("mamamamam"+e.getSQLState());
}
return "redirect:/owners/{ownerId}";
}
}
errors:
2018-11-19 23:26:24.025 WARN 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 1644, SQLState: 45000
2018-11-19 23:26:24.025 ERROR 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : Za duzo zwierzat
2018-11-19 23:26:24.040 ERROR 4532 --- [nio-8080-exec-6] o.a.c.c.C.[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path threw exception [Request processing failed; nested exception is org.springframework.orm.jpa.JpaSystemException: could not execute statement; nested exception is org.hibernate.exception.GenericJDBCException: could not execute statement] with root cause
java.sql.SQLException: Za duzo zwierzat
this is relevant part of mysql trigger (trigger works)
if (ilosc_zwierzat>1)
then
signal sqlstate '45000' SET MESSAGE_TEXT = 'Za duzo zwierzat';
end if;
I tried to do it this way but it doesn't work ( I've got a lot of text with errors on my page.)
@ExceptionHandler(SQLException.class)
public ModelAndView handleError(HttpServletRequest req, Exception ex) {
System.out.println("WIADMOSCcccccccccccccccccc"+ex.getMessage());
ModelAndView mav = new ModelAndView("error2");
mav.addObject("er", ex.getMessage());
mav.addObject("url", req.getRequestURL());
mav.setViewName("error2");
return mav;
}
java mysql spring
asked Nov 19 '18 at 22:37
wwwwwwww
234
add a comment |
I want to catch message of a trigger from mysql database and print it via my thymeleaft template but I came across a problem with catching message of that trigger
part of pet repository
void save(Pet pet) throws SQLException;
this is part of a controller, method of catching doesnt work and I don't know how to do it properly
@PostMapping("/pets/new")
public String processCreationForm(Owner owner, @Valid Pet pet, BindingResult result, ModelMap model) {
if (StringUtils.hasLength(pet.getName()) && pet.isNew() && owner.getPet(pet.getName(), true) != null){
result.rejectValue("name", "duplicate", "already exists");
}
owner.addPet(pet);
if (result.hasErrors()) {
model.put("pet", pet);
return VIEWS_PETS_CREATE_OR_UPDATE_FORM;
} else {
try {
this.pets.save(pet);
}catch (SQLException e){
System.out.println("mamamamam"+e.getSQLState());
}
return "redirect:/owners/{ownerId}";
}
}
errors:
2018-11-19 23:26:24.025 WARN 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 1644, SQLState: 45000
2018-11-19 23:26:24.025 ERROR 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : Za duzo zwierzat
2018-11-19 23:26:24.040 ERROR 4532 --- [nio-8080-exec-6] o.a.c.c.C.[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path threw exception [Request processing failed; nested exception is org.springframework.orm.jpa.JpaSystemException: could not execute statement; nested exception is org.hibernate.exception.GenericJDBCException: could not execute statement] with root cause
java.sql.SQLException: Za duzo zwierzat
this is relevant part of mysql trigger (trigger works)
if (ilosc_zwierzat>1)
then
signal sqlstate '45000' SET MESSAGE_TEXT = 'Za duzo zwierzat';
end if;
I tried to do it this way but it doesn't work ( I've got a lot of text with errors on my page.)
@ExceptionHandler(SQLException.class)
public ModelAndView handleError(HttpServletRequest req, Exception ex) {
System.out.println("WIADMOSCcccccccccccccccccc"+ex.getMessage());
ModelAndView mav = new ModelAndView("error2");
mav.addObject("er", ex.getMessage());
mav.addObject("url", req.getRequestURL());
mav.setViewName("error2");
return mav;
}
java mysql spring
asked Nov 19 '18 at 22:37
wwwwwwww
234
Evidently the posted code doesn't catch the exception, the stacktrace doesn't seem to list the controller. is this using open-session-in-view? where are the transaction boundaries? remember Hibernate uses transactional write-behind so the SQL doesn't fire until a flush or commit happens. Consider using services, see Service layer and controller: who takes care of what?
– Nathan Hughes
Nov 19 '18 at 22:50
add a comment |
I want to catch message of a trigger from mysql database and print it via my thymeleaft template but I came across a problem with catching message of that trigger
part of pet repository
void save(Pet pet) throws SQLException;
this is part of a controller, method of catching doesnt work and I don't know how to do it properly
@PostMapping("/pets/new")
public String processCreationForm(Owner owner, @Valid Pet pet, BindingResult result, ModelMap model) {
if (StringUtils.hasLength(pet.getName()) && pet.isNew() && owner.getPet(pet.getName(), true) != null){
result.rejectValue("name", "duplicate", "already exists");
}
owner.addPet(pet);
if (result.hasErrors()) {
model.put("pet", pet);
return VIEWS_PETS_CREATE_OR_UPDATE_FORM;
} else {
try {
this.pets.save(pet);
}catch (SQLException e){
System.out.println("mamamamam"+e.getSQLState());
}
return "redirect:/owners/{ownerId}";
}
}
errors:
2018-11-19 23:26:24.025 WARN 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 1644, SQLState: 45000
2018-11-19 23:26:24.025 ERROR 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : Za duzo zwierzat
2018-11-19 23:26:24.040 ERROR 4532 --- [nio-8080-exec-6] o.a.c.c.C.[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path threw exception [Request processing failed; nested exception is org.springframework.orm.jpa.JpaSystemException: could not execute statement; nested exception is org.hibernate.exception.GenericJDBCException: could not execute statement] with root cause
java.sql.SQLException: Za duzo zwierzat
this is relevant part of mysql trigger (trigger works)
if (ilosc_zwierzat>1)
then
signal sqlstate '45000' SET MESSAGE_TEXT = 'Za duzo zwierzat';
end if;
I tried to do it this way but it doesn't work ( I've got a lot of text with errors on my page.)
@ExceptionHandler(SQLException.class)
public ModelAndView handleError(HttpServletRequest req, Exception ex) {
System.out.println("WIADMOSCcccccccccccccccccc"+ex.getMessage());
ModelAndView mav = new ModelAndView("error2");
mav.addObject("er", ex.getMessage());
mav.addObject("url", req.getRequestURL());
mav.setViewName("error2");
return mav;
}
java mysql spring
asked Nov 19 '18 at 22:37
wwwwwwww
234
I want to catch message of a trigger from mysql database and print it via my thymeleaft template but I came across a problem with catching message of that trigger
part of pet repository
void save(Pet pet) throws SQLException;
this is part of a controller, method of catching doesnt work and I don't know how to do it properly
@PostMapping("/pets/new")
public String processCreationForm(Owner owner, @Valid Pet pet, BindingResult result, ModelMap model) {
if (StringUtils.hasLength(pet.getName()) && pet.isNew() && owner.getPet(pet.getName(), true) != null){
result.rejectValue("name", "duplicate", "already exists");
}
owner.addPet(pet);
if (result.hasErrors()) {
model.put("pet", pet);
return VIEWS_PETS_CREATE_OR_UPDATE_FORM;
} else {
try {
this.pets.save(pet);
}catch (SQLException e){
System.out.println("mamamamam"+e.getSQLState());
}
return "redirect:/owners/{ownerId}";
}
}
errors:
2018-11-19 23:26:24.025 WARN 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : SQL Error: 1644, SQLState: 45000
2018-11-19 23:26:24.025 ERROR 4532 --- [nio-8080-exec-6] o.h.engine.jdbc.spi.SqlExceptionHelper : Za duzo zwierzat
2018-11-19 23:26:24.040 ERROR 4532 --- [nio-8080-exec-6] o.a.c.c.C.[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path threw exception [Request processing failed; nested exception is org.springframework.orm.jpa.JpaSystemException: could not execute statement; nested exception is org.hibernate.exception.GenericJDBCException: could not execute statement] with root cause
java.sql.SQLException: Za duzo zwierzat
this is relevant part of mysql trigger (trigger works)
if (ilosc_zwierzat>1)
then
signal sqlstate '45000' SET MESSAGE_TEXT = 'Za duzo zwierzat';
end if;
I tried to do it this way but it doesn't work ( I've got a lot of text with errors on my page.)
@ExceptionHandler(SQLException.class)
public ModelAndView handleError(HttpServletRequest req, Exception ex) {
System.out.println("WIADMOSCcccccccccccccccccc"+ex.getMessage());
ModelAndView mav = new ModelAndView("error2");
mav.addObject("er", ex.getMessage());
mav.addObject("url", req.getRequestURL());
mav.setViewName("error2");
return mav;
}
java mysql spring
java mysql spring
asked Nov 19 '18 at 22:37
wwwwwwww
234
asked Nov 19 '18 at 22:37
wwwwwwww
234
edited Nov 20 '18 at 1:08
wwww
asked Nov 19 '18 at 22:37
wwwwwwww
234
asked Nov 19 '18 at 22:37
wwwwwwww
234
asked Nov 19 '18 at 22:37
wwwwwwww
234
234
Evidently the posted code doesn't catch the exception, the stacktrace doesn't seem to list the controller. is this using open-session-in-view? where are the transaction boundaries? remember Hibernate uses transactional write-behind so the SQL doesn't fire until a flush or commit happens. Consider using services, see Service layer and controller: who takes care of what?
– Nathan Hughes
Nov 19 '18 at 22:50
add a comment |
Evidently the posted code doesn't catch the exception, the stacktrace doesn't seem to list the controller. is this using open-session-in-view? where are the transaction boundaries? remember Hibernate uses transactional write-behind so the SQL doesn't fire until a flush or commit happens. Consider using services, see Service layer and controller: who takes care of what?
– Nathan Hughes
Nov 19 '18 at 22:50
Evidently the posted code doesn't catch the exception, the stacktrace doesn't seem to list the controller. is this using open-session-in-view? where are the transaction boundaries? remember Hibernate uses transactional write-behind so the SQL doesn't fire until a flush or commit happens. Consider using services, see Service layer and controller: who takes care of what?
– Nathan Hughes
Nov 19 '18 at 22:50
Evidently the posted code doesn't catch the exception, the stacktrace doesn't seem to list the controller. is this using open-session-in-view? where are the transaction boundaries? remember Hibernate uses transactional write-behind so the SQL doesn't fire until a flush or commit happens. Consider using services, see Service layer and controller: who takes care of what?
– Nathan Hughes
Nov 19 '18 at 22:50
add a comment |
1 Answer
1
active
oldest
votes
You can write your custom exception handling
Using @ControllerAdvice which is global exception handling
Or you can use @ExceptionHandler which applies only inside controller you defined
You can refer to this guide:
https://spring.io/blog/2013/11/01/exception-handling-in-spring-mvc
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can write your custom exception handling
Using @ControllerAdvice which is global exception handling
Or you can use @ExceptionHandler which applies only inside controller you defined
You can refer to this guide:
https://spring.io/blog/2013/11/01/exception-handling-in-spring-mvc
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
add a comment |
You can write your custom exception handling
Using @ControllerAdvice which is global exception handling
Or you can use @ExceptionHandler which applies only inside controller you defined
You can refer to this guide:
https://spring.io/blog/2013/11/01/exception-handling-in-spring-mvc
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
add a comment |
You can write your custom exception handling
Using @ControllerAdvice which is global exception handling
Or you can use @ExceptionHandler which applies only inside controller you defined
You can refer to this guide:
https://spring.io/blog/2013/11/01/exception-handling-in-spring-mvc
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
You can write your custom exception handling
Using @ControllerAdvice which is global exception handling
Or you can use @ExceptionHandler which applies only inside controller you defined
You can refer to this guide:
https://spring.io/blog/2013/11/01/exception-handling-in-spring-mvc
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
edited Nov 19 '18 at 23:20
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
answered Nov 19 '18 at 22:40
Mykhailo MoskuraMykhailo Moskura
839113
839113
add a comment |
add a comment |
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Evidently the posted code doesn't catch the exception, the stacktrace doesn't seem to list the controller. is this using open-session-in-view? where are the transaction boundaries? remember Hibernate uses transactional write-behind so the SQL doesn't fire until a flush or commit happens. Consider using services, see Service layer and controller: who takes care of what?
– Nathan Hughes
Nov 19 '18 at 22:50