我正在使用Query DSL并希望我的结果集返回一个Page Object。在Query DSL中有没有办法做到这一点?如果是这样我的查询是什么样的? 我正在使用JPAQuery,我有我的QClasses
方法结构就是这个
public Page<Object> searchPerson(String name,String phone){
Page<Object> results=null;
JPQLQuery query = new JPAQuery(entityManager);
QPerson person = QPerson.person;
//I am assuming my query would go here
results = query.from(person). ?????
return results;}
帮助!
答案 0 :(得分:2)
以下是我使用QueryDSL的Paging
实现。 PageRequest
定义查询的参数(限制和页面):
public class PageRequest {
protected Long page = 1l;// 1 is the first page
protected Integer limit = 10;
public PageRequest(Long page, Integer limit) {
this.limit = limit;
this.page = page;
}
public Long getPage() {
return page;
}
public Integer getLimit() {
return limit;
}
public Long getOffset() {
return (page - 1l) * limit;
}
}
Page
类包含查询的结果(此处为属性objects
),并且可以实现创建好的分页链接的方法。
public class Page<T> extends PageRequest {
protected Collection<T> objects;
private Long totalCount;
private Long pageCount;
private Boolean hasPageLinkPrev;
private Boolean hasPageLinkNext;
private Collection<Long> pageLinks;
public Page(Long page, Integer limit, Long totalCount, Collection<T> objects) {
this.page = page;
this.limit = limit;
this.totalCount = totalCount;
this.objects = objects;
this.pageCount = totalCount / limit;
if (totalCount % limit > 0) {
this.pageCount = this.pageCount + 1;
}
this.hasPageLinkPrev = page > 1;
this.hasPageLinkNext = page < this.pageCount;
this.pageLinks = new ArrayList<>();
if (this.pageCount != 1) {
this.pageLinks.add(1l);
if (page > 3l) {
this.pageLinks.add(-1l);
}
if (page > 2l) {
if (page.equals(this.pageCount) && this.pageCount > 3l) {
this.pageLinks.add(page - 2l);
}
this.pageLinks.add(page - 1l);
}
if (page != 1l && !page.equals(this.pageCount)) {
this.pageLinks.add(page);
}
if (page < this.pageCount - 1l) {
this.pageLinks.add(page + 1l);
if (page == 1l && this.pageCount > 3l) {
this.pageLinks.add(page + 2l);
}
}
if (page < this.pageCount - 2l) {
this.pageLinks.add(-1l);
}
this.pageLinks.add(this.pageCount);
}
}
public Page(PageRequest pageRequest, Long totalCount, Collection<T> objects) {
this(pageRequest.getPage(), pageRequest.getLimit(), totalCount, objects);
}
public Long getTotalCount() {
return this.totalCount;
}
public Long getPageCount() {
return this.pageCount;
}
public Long getPage() {
return this.page;
}
public Integer getLimit() {
return this.limit;
}
public Boolean getHasPageLinkPrev() {
return this.hasPageLinkPrev;
}
public Boolean getHasPageLinkNext() {
return hasPageLinkNext;
}
public Collection<Long> getPageLinks() {
return pageLinks;
}
public Collection<T> getObjects() {
return objects;
}
}
使用该stuf创建查询并将结果放入页面对象并不是很困难。一种可能性是在存储库类的基类中编写泛型方法:
protected <T> Page<T> getPage(JPQLQuery<T> query, PageRequest pageRequest) {
List<T> resultList = query
.offset(pageRequest.getOffset())
.limit(pageRequest.getLimit())
.fetch();
Long totalCount = query.fetchCount();
return new Page<T>(pageRequest, totalCount, resultList);
}
在您的存储库类中,您可以为特定用例创建查询。然后,您可以使用getPage
方法将结果生成为Page
。
public Page<Person> searchPerson(String name,
String phone,
PageRequest request){
Page<Person> results=null;
JPQLQuery<Person> query = new JPAQuery<>(entityManager);
QPerson person = QPerson.person;
query = query.from(person)
.where(person.name.eq(name)
.and(person.phone.eq(phone)));
return getPage(query, request);
}
答案 1 :(得分:0)
上面的解决方案是使用在上面的方法上实现的BooleanBuilder并更改方法名称以返回Person Object。 请检查BooleanBuilder
QPerson person= QPerson.person;
BooleanBuilder builder = this.getBuilder(name, phone,page, pageSize, sortFlag, sortItem);
PageRequest pg = getPRequest(page, pageSize);
Page<Person> pages personRepo.findAll(builder,pg);
return pages;
然后实现getBuilder方法,它是下面的
public BooleanBuilder getBuilder(String name, String phone, Integer page, Integer pageSize, String sortFlag, String sortItem) {
QPerson person = QPerson.person;
BooleanBuilder builder = new BooleanBuilder();
builder.and(person.name.startsWith(name));
return builder;
}
最后将getPRequest方法实现为以下
public PageRequest getPRequest(Integer page, Integer pageSize) {
return new PageRequest(page, pageSize);
}
Oooooh Happy Days!