选择
(select count(*) from table2 where table2.table2Seq = table1.table1Seq) as count1,
(select count(*) from table2 where table2.table2Seq = table1.table1Seq and table2Yn = true) as count2,
(select count(*) from table2 where table2.table2Seq = table1.table1Seq) / (select count(*) from table2 where table2.table2Seq = table1.table1Seq and table2Yn = true) as count3
从…
table1
排序依据
count3
**我想要这个查询。你是怎么把它变成queryDSL的?我想对子查询中的计数器除以子查询中的计数器进行排序。**
NumberPath<Long> count1 = Expressions.numberPath(Long.class, “count1”);
NumberPath<Long> count2 = Expressions.numberPath(Long.class, “count2”);
NumberPath<Long> count3 = Expressions.numberPath(Long.class, “count3”);
JPQLQuery<OnlineTrainingCourseDto> query = getQuerydsl().createQuery()
.select(
new Dto(
ExpressionUtils.as(
JPAExpressions.select(table2.table2Seq.count())
.from(table2)
.where(table2.table2Seq.eq(table1.table1Seq)),
count1
),
ExpressionUtils.as(
JPAExpressions.select(table2.table2Seq.count())
.from(table2)
.where(table2.table2Seq.eq(table1.table1Seq)
.and(table2.table2Yn.eq(true))),
count2
),
***????? count1 / count2 as count3 ?????***
)
)
)
.from(table1)
发布于 2020-08-24 15:55:54
可以将表达式转换为数字表达式以访问除法运算符:
JPAExpressions.asNumber(
JPAExpressions.select(table2.table2Seq.count())
.from(table2)
.where(table2.table2Seq.eq(table1.table1Seq)))
.divide(
JPAExpressions.asNumber(JPAExpressions.select(table2.table2Seq.count())
.from(table2)
.where(table2.table2Seq.eq(table1.table1Seq)
.and(table2.table2Yn.eq(true))))
).as("count3")
https://stackoverflow.com/questions/63554873
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