JPAQuery 与 Spring Pageable

Question

我想使用org.springframework.data.domain.Pageablewithcom.querydsl.jpa.impl.JPAQuery有没有办法或者有可靠的解决方法如何使 Pageable 与 JPAQuery 一起工作？

Answer 1

所以我发现为什么要让他们一起工作发布完整的示例：

import com.querydsl.core.BooleanBuilder;
import com.querydsl.core.types.Order;
import com.querydsl.core.types.OrderSpecifier;
import com.querydsl.core.types.dsl.BooleanExpression;
import com.querydsl.core.types.dsl.PathBuilder;
import com.querydsl.jpa.impl.JPAQuery;
import org.springframework.data.domain.Pageable;
import org.springframework.data.domain.Sort;
import javax.persistence.EntityManager;

public class Example {
    public void example(Pageable pageable, EntityManager em){
        QCustomer qCustomer = QCustomer.customer;

        JPAQuery<?> query = new JPAQuery<>(em);
        query.from(qCustomer);

        BooleanBuilder builder = new BooleanBuilder();
        builder.and(qCustomer.email.likeIgnoreCase("c@m.c"));

        JPAQuery<?> where = query.where(builder);
        // This for loop is making it work with Pagable
        query.offset(pageable.getOffset());
        query.limit(pageable.getPageSize());
        PathBuilder<Customer> entityPath = new PathBuilder<>(Customer.class, "customer");
        for (Sort.Order order : pageable.getSort()) {
            PathBuilder<Object> path = entityPath.get(order.getProperty());
            query.orderBy(new OrderSpecifier(Order.valueOf(order.getDirection().name()), path));
        }

        List<Customer> resultList = query.createQuery().getResultList();
    }
}