Spring Data JPA Specification with MayToMany relation
0
I'm trying to use Specification to build a search criteria with Spring Data JPA.
I have 2 entities
@Entity
public class User {
@Id
private Long userId;
@ManyToMany
@JoinColumn(name="locationId")
@JsonIgnore
private List<Location> locations;
//getters and setters
And Location
public class Location {
@Id
private long locationId;
@ManyToMany(fetch = FetchType.LAZY, cascade = { CascadeType.PERSIST, CascadeType.REMOVE, CascadeType.DETACH,
CascadeType.REFRESH }, mappedBy = "locations")
@JsonIgnore
private List<User> users;
//getters and setters
Now I would like to find all users with a given list of locations, so I'm trying to build the predicate.
My problem is getting the relation table between the 2 entities... I'm trying in this way
Root<Location> second = query.from(Location.class);
Expression<Collection<Root>> userLocations = second.get("users");
return builder.and(builder.isMember(root, userLocations));
This is the sql query i need
select distinct * from User u
join User_Location ul on u.userId = ul.`users_userId`
where ul.`locations_locationId` in (1,2,3...)
The main difficult for me is telling Spring to use the relation table User_Location which is not an entity...
But I indeed miss something...
java jpa spring-data-jpa many-to-many criteria-api
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
add a comment |
0
I'm trying to use Specification to build a search criteria with Spring Data JPA.
I have 2 entities
@Entity
public class User {
@Id
private Long userId;
@ManyToMany
@JoinColumn(name="locationId")
@JsonIgnore
private List<Location> locations;
//getters and setters
And Location
public class Location {
@Id
private long locationId;
@ManyToMany(fetch = FetchType.LAZY, cascade = { CascadeType.PERSIST, CascadeType.REMOVE, CascadeType.DETACH,
CascadeType.REFRESH }, mappedBy = "locations")
@JsonIgnore
private List<User> users;
//getters and setters
Now I would like to find all users with a given list of locations, so I'm trying to build the predicate.
My problem is getting the relation table between the 2 entities... I'm trying in this way
Root<Location> second = query.from(Location.class);
Expression<Collection<Root>> userLocations = second.get("users");
return builder.and(builder.isMember(root, userLocations));
This is the sql query i need
select distinct * from User u
join User_Location ul on u.userId = ul.`users_userId`
where ul.`locations_locationId` in (1,2,3...)
The main difficult for me is telling Spring to use the relation table User_Location which is not an entity...
But I indeed miss something...
java jpa spring-data-jpa many-to-many criteria-api
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
You can break down a bidirectional Many-to-Many relationship to two One-To-Many relationships. The config can be found here.
– huytmb
Nov 13 '18 at 8:52
1
You don't need to use the join table in your query. Simply do "select distrinct u from User u where u.locations in (l1, l2, l3, …)" In criteriaBuilder the class is a "Join" class, so Join<User, Location> locations = root.join("locations)" if I remember correctly. Then "query.where( builder.in(locations, locationList))". Something along those lines.
– K.Nicholas
Nov 13 '18 at 20:00
thanks, this works perfectly
– besmart
Nov 14 '18 at 13:01
add a comment |
0
0
0
I'm trying to use Specification to build a search criteria with Spring Data JPA.
I have 2 entities
@Entity
public class User {
@Id
private Long userId;
@ManyToMany
@JoinColumn(name="locationId")
@JsonIgnore
private List<Location> locations;
//getters and setters
And Location
public class Location {
@Id
private long locationId;
@ManyToMany(fetch = FetchType.LAZY, cascade = { CascadeType.PERSIST, CascadeType.REMOVE, CascadeType.DETACH,
CascadeType.REFRESH }, mappedBy = "locations")
@JsonIgnore
private List<User> users;
//getters and setters
Now I would like to find all users with a given list of locations, so I'm trying to build the predicate.
My problem is getting the relation table between the 2 entities... I'm trying in this way
Root<Location> second = query.from(Location.class);
Expression<Collection<Root>> userLocations = second.get("users");
return builder.and(builder.isMember(root, userLocations));
This is the sql query i need
select distinct * from User u
join User_Location ul on u.userId = ul.`users_userId`
where ul.`locations_locationId` in (1,2,3...)
The main difficult for me is telling Spring to use the relation table User_Location which is not an entity...
But I indeed miss something...
java jpa spring-data-jpa many-to-many criteria-api
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
I'm trying to use Specification to build a search criteria with Spring Data JPA.
I have 2 entities
@Entity
public class User {
@Id
private Long userId;
@ManyToMany
@JoinColumn(name="locationId")
@JsonIgnore
private List<Location> locations;
//getters and setters
And Location
public class Location {
@Id
private long locationId;
@ManyToMany(fetch = FetchType.LAZY, cascade = { CascadeType.PERSIST, CascadeType.REMOVE, CascadeType.DETACH,
CascadeType.REFRESH }, mappedBy = "locations")
@JsonIgnore
private List<User> users;
//getters and setters
Now I would like to find all users with a given list of locations, so I'm trying to build the predicate.
My problem is getting the relation table between the 2 entities... I'm trying in this way
Root<Location> second = query.from(Location.class);
Expression<Collection<Root>> userLocations = second.get("users");
return builder.and(builder.isMember(root, userLocations));
This is the sql query i need
select distinct * from User u
join User_Location ul on u.userId = ul.`users_userId`
where ul.`locations_locationId` in (1,2,3...)
The main difficult for me is telling Spring to use the relation table User_Location which is not an entity...
But I indeed miss something...
java jpa spring-data-jpa many-to-many criteria-api
java jpa spring-data-jpa many-to-many criteria-api
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
edited Nov 13 '18 at 8:28
besmart
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
asked Nov 12 '18 at 21:45
besmartbesmart
65522455
65522455
You can break down a bidirectional Many-to-Many relationship to two One-To-Many relationships. The config can be found here.
– huytmb
Nov 13 '18 at 8:52
1
You don't need to use the join table in your query. Simply do "select distrinct u from User u where u.locations in (l1, l2, l3, …)" In criteriaBuilder the class is a "Join" class, so Join<User, Location> locations = root.join("locations)" if I remember correctly. Then "query.where( builder.in(locations, locationList))". Something along those lines.
– K.Nicholas
Nov 13 '18 at 20:00
thanks, this works perfectly
– besmart
Nov 14 '18 at 13:01
add a comment |
You can break down a bidirectional Many-to-Many relationship to two One-To-Many relationships. The config can be found here.
– huytmb
Nov 13 '18 at 8:52
1
You don't need to use the join table in your query. Simply do "select distrinct u from User u where u.locations in (l1, l2, l3, …)" In criteriaBuilder the class is a "Join" class, so Join<User, Location> locations = root.join("locations)" if I remember correctly. Then "query.where( builder.in(locations, locationList))". Something along those lines.
– K.Nicholas
Nov 13 '18 at 20:00
thanks, this works perfectly
– besmart
Nov 14 '18 at 13:01
You can break down a bidirectional Many-to-Many relationship to two One-To-Many relationships. The config can be found here.
– huytmb
Nov 13 '18 at 8:52
You can break down a bidirectional Many-to-Many relationship to two One-To-Many relationships. The config can be found here.
– huytmb
Nov 13 '18 at 8:52
1
1
You don't need to use the join table in your query. Simply do "select distrinct u from User u where u.locations in (l1, l2, l3, …)" In criteriaBuilder the class is a "Join" class, so Join<User, Location> locations = root.join("locations)" if I remember correctly. Then "query.where( builder.in(locations, locationList))". Something along those lines.
– K.Nicholas
Nov 13 '18 at 20:00
You don't need to use the join table in your query. Simply do "select distrinct u from User u where u.locations in (l1, l2, l3, …)" In criteriaBuilder the class is a "Join" class, so Join<User, Location> locations = root.join("locations)" if I remember correctly. Then "query.where( builder.in(locations, locationList))". Something along those lines.
– K.Nicholas
Nov 13 '18 at 20:00
thanks, this works perfectly
– besmart
Nov 14 '18 at 13:01
thanks, this works perfectly
– besmart
Nov 14 '18 at 13:01
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53270564%2fspring-data-jpa-specification-with-maytomany-relation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53270564%2fspring-data-jpa-specification-with-maytomany-relation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You can break down a bidirectional Many-to-Many relationship to two One-To-Many relationships. The config can be found here.
– huytmb
Nov 13 '18 at 8:52
1
You don't need to use the join table in your query. Simply do "select distrinct u from User u where u.locations in (l1, l2, l3, …)" In criteriaBuilder the class is a "Join" class, so Join<User, Location> locations = root.join("locations)" if I remember correctly. Then "query.where( builder.in(locations, locationList))". Something along those lines.
– K.Nicholas
Nov 13 '18 at 20:00
thanks, this works perfectly
– besmart
Nov 14 '18 at 13:01