how to catch sqlException or signal message from mysql trigger











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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;
}









share|improve this question
























  • 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 at 22:50

















up vote
1
down vote

favorite












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;
}









share|improve this question
























  • 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 at 22:50















up vote
1
down vote

favorite









up vote
1
down vote

favorite











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;
}









share|improve this question















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






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share|improve this question








edited Nov 20 at 1:08

























asked Nov 19 at 22:37









wwww

224




224












  • 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 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 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 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 at 22:50














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up vote
0
down vote













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






share|improve this answer























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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    0
    down vote













    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






    share|improve this answer



























      up vote
      0
      down vote













      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






      share|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        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






        share|improve this answer














        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







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 19 at 23:20

























        answered Nov 19 at 22:40









        Mykhailo Moskura

        804112




        804112






























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