通过查找table2的center_geom与表3中的home_location之间的距离来更新表1“distance”列中的值。
表1
obu_id end_location trip_id end_loc_adj distance
1 51 1234
2 57 1357 44
2 63 1351
3 21 1212 20
3 23 4313
表2
id obu_id center_geom
int int geom
51 2 "0101000020ED0800006DFFCAA2A2553341B20E4717774E0C41"
52 3 "0101000020ED080000AE47E17A35F73341FE65F764723C0841"
57 3 "0101000020ED0800006DFFCAA2A2553341B20E4717774E0C41"
21 4 "0101000020ED080000B81E852BC555334186048C9EB1C21141"
表3
hhid obu_id home_location
(character Int geometry
varying)
11 1 "0101000020ED08000082E2C7A0B2413341BC5818A21F000941"
15 2 "0101000020ED080000B81E852BC555334186048C9EB1C21141"
17 3 "0101000020ED0800006DFFCAA2A2553341B20E4717774E0C41"
17 4 "0101000020ED080000B81E852BC555334186048C9EB1C21141"
22 7 "0101000020ED080000AE47E17A35F73341FE65F764723C0841"
我尝试了以下内容:
select ST_Distance_Sphere ( ST_AsText(cast(v.house_location AS text)) , ST_AsText(cast(l.center AS text)))
来自public.locations l,clean.vehicles v 其中l.obu_id = v.obu_id 和l.obu_id = 3
但这适用于单个obu_id
我想用一个函数一次完成它。 请告诉我如何做到这一点。
感谢。
答案 0 :(得分:1)
house_loc:=(select house_location from table1 where obu_id=b.obu_id);
house_loc_geom := ST_Transform(house_loc,4269);
IF b.end_location_adj IS null THEN
end_loc:= (select center from table2 where id=b.end_location and obu_id= b.obu_id);
else
end_loc:= (select center from tabl2 where id = b.end_location_adj and obu_id = b.obu_id);
end if;
center_geom:=ST_Transform(end_loc,4269);
UPDATE table 1 set dist_from_home_in_meter=distance
where obu_id=b.obu_id and trip_id=b.trip_id and end_location=b.end_location;
这是接近回答,需要一些修改。