将SQL查询(使用子查询连接)转换为KnexJS
将SQL查询(使用子查询连接)转换为KnexJS
提问于 2018-12-19 13:14:03
我正在尝试将SQL查询转换为KnexJS格式，但当前的KnexJS查询显示以下错误。
- 在"as“堆栈或附近出现语法错误: error:在"as"
或附近出现语法错误
以下是原始查询，也是我一直在为KnexJS处理的查询。请更正我的KnexJS查询。基本上，我想知道如何构建KnexJS查询-内部连接子查询提前谢谢！
原始SQL查询：
select DATE_RANGE.START_DATE, DATE_RANGE.END_DATE, count (distinct DATE) as DATE_COUNT
from TASK_HISTORY
join
(select
STORE_ID,
to_number(to_char(to_date(to_char(DATE,'99999999'),'YYYYMMDD') - 1,'YYYYMMDD'),'99999999') as END_DATE
, count (distinct DATE) as REC_COUNT
, to_number(to_char(to_date(to_char(lag (DATE) over (order by DATE asc),'99999999'),'YYYYMMDD') + 1,'YYYYMMDD'),'99999999') as START_DATE
, count (case when FINISH_TIME is not null then 1 end) as COUNT_FINISHED
, count (case when FINISH_TIME is null then 1 end) as COUNT_UNFINISHED
from TASK_HISTORY
where STORE_ID = 43
group by DATE, STORE_ID
having count (case when FINISH_TIME is not null then 1 end) = 0
order by DATE)
as DATE_RANGE
on TASK_HISTORY.DATE >= DATE_RANGE.START_DATE
AND TASK_HISTORY.DATE <= DATE_RANGE.END_DATE
AND TASK_HISTORY.STORE_ID = 43
group by DATE_RANGE.START_DATE, DATE_RANGE.END_DATE, DATE_RANGE.REC_COUNT
order by DATE_COUNT desc, START_DATE desc更新：
以下是对我有效的解决方案：
await db
.table("task_history")
.select('date_range.start_date', 'date_range.end_date')
.select(db.raw(`count(distinct date) as date_count`))
.join(
db
.select('task_history.store_id')
.table('task_history')
.select(db.raw(
`to_number(to_char(to_date(to_char(date,'99999999'),'YYYYMMDD') - 1,'YYYYMMDD'),'99999999') as end_date`
))
.select(db.raw(`count(distinct date) as rec_count`))
.select(db.raw(
`to_number(to_char(to_date(to_char(lag (date) over (order by date asc),'99999999'),'YYYYMMDD') + 1,'YYYYMMDD'),'99999999') as start_date`
))
.select(db.raw(`count(case when FINISH_TIME is not null then 1 end) as COUNT_FINISHED`))
.select(db.raw(`count(case when FINISH_TIME is null then 1 end) as COUNT_UNFINISHED`))
.where('task_history.store_id', 43)
.groupBy('task_history.date', 'task_history.store_id')
.having(db.raw(`count(case when FINISH_TIME is not null then 1 end) = 0 order by date`))
.as('date_range'),
function () {
this.on('task_history.date', '>=', 'date_range.start_date')
.andOn('task_history.date', '<=', 'date_range.end_date')
.andOn('task_history.store_id', 43)
}
)
.groupBy('date_range.start_date', 'date_range.end_date', 'date_range.rec_count')
.orderBy('date_count', 'desc')
.orderBy('start_date', 'desc')回答 1
Stack Overflow用户
回答已采纳
发布于 2018-12-19 23:14:38
这可能会对你有帮助-
const sql = db.table("task_history")
.select('DATE_RANGE.START_DATE', 'DATE_RANGE.END_DATE')
.select(db.raw(`count(distinct DATE) as DATE_COUNT`))
.innerJoin(
db.select('store_id')
.table('task_history')
.select(db.raw(
`to_number(to_char(to_date(to_char(DATE,'99999999'),'YYYYMMDD') - 1,'YYYYMMDD'),'99999999') as END_DATE`
))
.select(db.raw(`count(distinct DATE) as REC_COUNT`))
.select(db.raw(
`to_number(to_char(to_date(to_char(lag (DATE) over (order by DATE asc),'99999999'),'YYYYMMDD') + 1,'YYYYMMDD'),'99999999') as START_DATE`
))
.select(db.raw(`count(case when FINISH_TIME is not null then 1 end) as COUNT_FINISHED`))
.select(db.raw(`count(case when FINISH_TIME is null then 1 end) as COUNT_UNFINISHED`))
.where('store_id', 43)
.groupBy('date', 'store_id')
.having(db.raw(`count(case when FINISH_TIME is not null then 1 end) = 0 order by DATE`))
.as('DATE_RANGE')
, function () {
this.on('DATE_RANGE.START_DATE', '>=', 'TASK_HISTORY.DATE')
.andOn('TASK_HISTORY.DATE', '<=', 'DATE_RANGE.END_DATE')
.andOn('TASK_HISTORY.STORE_ID', 43)
})
.where('task_history.date', '>=', 'DATE_RANGE.START_DATE')
.where('task_history.date', '<=', 'DATE_RANGE.END_DATE')
.groupBy('DATE_RANGE.START_DATE', 'DATE_RANGE.END_DATE', 'DATE_RANGE.REC_COUNT')
.orderBy('DATE_COUNT', 'desc')
.orderBy('START_DATE', 'desc')
.toSQL();
console.log(sql);页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/53844960
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