我有一个jooq生成的查询的性能问题,它应该等于普通的字符串查询。 jooq查询如下所示:
return ctx.select(AT_TRAIL.POINT)
.from(AT_TRAIL)
.where(AT_TRAIL.ID.le(
ctx.select(AT_TRAIL.ID)
.from(
ctx.select(AT_TRAIL.ID, AT_TRAIL.POINT, field(
"point <-> ( " +
" select point " +
" from at_shelter " +
" where id = ( " +
" select at_shelter " +
" from at_last_shelter " +
" ) "+
")"
).as("dist"))
.from(AT_TRAIL)
.orderBy(field("dist").asc())
.limit(1)
)
))
.orderBy(AT_TRAIL.ID.asc())
.fetch()
.map(r -> {
PGpoint point = r.get(AT_TRAIL.POINT, PGpoint.class);
return ImmutableMap.of("lat", point.x, "lng", point.y);
});
我的普通字符串查询看起来像这样
return ctx.fetch(
" select point " +
" from at_trail " +
" where id <= ( " +
" select id " +
" from ( " +
" select id, point, point <-> ( " +
" select point " +
" from at_shelter " +
" where id = ( " +
" select at_shelter " +
" from at_last_shelter " +
" ) " +
" ) as dist " +
" from at_trail " +
" order by dist asc " +
" limit 1 " +
" ) t " +
" ) " +
"order by id asc"
)
.map(r -> {
PGpoint point = r.get(AT_TRAIL.POINT, PGpoint.class);
return ImmutableMap.of("lat", point.x, "lng", point.y);
});
我将jooq生成的查询与另一个查询进行了比较。它们在表别名中有所不同。 jooq生成as "alias_108340908"
,而我只使用t
。并且jooq完全定义了"public"."at_trail"."point"
之类的列名和表。否则两个查询是相同的。但是,使用jooq生成的查询最多需要30秒才能完成,而另一个只需要几毫秒。
导致性能问题的原因是什么?资格?如何禁用它/加快查询速度?
答案 0 :(得分:2)
您的jOOQ查询错误(假设您的纯SQL查询是正确的)。考虑一下:
return ctx.select(AT_TRAIL.POINT)
.from(AT_TRAIL)
.where(AT_TRAIL.ID.le(
ctx.select(AT_TRAIL.ID) // This is the outer query's ID, not the local ID
.from(...)
))
.orderBy(AT_TRAIL.ID.asc())
.fetch()
你打算写的是:
return ctx.select(AT_TRAIL.POINT)
.from(AT_TRAIL)
.where(AT_TRAIL.ID.le(
ctx.select(field("id", AT_TRAIL.ID.getDataType())) // Better
.from(...)
))
.orderBy(AT_TRAIL.ID.asc())
.fetch()
现在,您当然可以简化原始查询,以简化此操作。例如。这似乎做了同样的事情:
dist
从SELECT
移至ORDER BY
,删除一个嵌套查询:select point
from at_trail
where id <= (
select id
from at_trail
order by point <-> (
select point
from at_shelter
where id = (
select at_shelter
from at_last_shelter
)
) asc
limit 1
)
order by id asc
以上查询将如下:
ctx.select(AT_TRAIL.POINT)
.from(AT_TRAIL)
.where(AT_TRAIL.ID.le(
select(AT_TRAIL.ID) // Now, no scoping problem anymore
.from(AT_TRAIL)
.orderBy(field("{0} <-> {1}", // jOOQ doesn't support this op, resorting to plain SQL
AT_TRAIL.POINT,
select(AT_SHELTER.POINT)
.from(AT_SHELTER)
.where(AT_SHELTER.ID.eq(
select(AT_LAST_SHELTER.AT_SHELTER)
.from(AT_LAST_SHELTER)
))
).asc())
.limit(1)
))
.orderBy(AT_TRAIL.ID.asc())
.fetch();
取决于你正在做什么(我把它读作直到离最后一个避难所最近点的路径),这可能更加优化,但为了这个问题,我认为这是已经很不错了。