所以，我必须编写一个查询，其中给定一堆带(x, y, z)坐标的点，我需要为给定{{1}列表中的每个点找到最接近的place }}第
place坐标是用户提供的，因此，数据库中不存在，我使用x, y, z作为临时表的一种工作。
SELECT... UNION的列表实际上是我place id places_table的{{1}}，4030764,4030734,4030752,3948...看起来像 -
places_table性能 真的 至关重要，因此我尝试计算所有的平方距离（每个点对位置之间）以及选择每个点的最小距离位置一气呵成。
我试图写这个查询，（可能不是一个非常好的尝试......）
CREATE TABLE `places_table` (
`place_id` int(10) unsigned NOT NULL,
`latlon_id` int(11) DEFAULT NULL,
`latitude` double DEFAULT NULL,
`longitude` double DEFAULT NULL,
`x` double DEFAULT NULL,
`y` double DEFAULT NULL,
`z` double DEFAULT NULL,
)
它会起作用但SELECT point.index, (@px:=point.x) AS point_x, (@py:=point.y) AS point_y,
(@pz:=point.z) AS point_z, place.lat_lon_id, place.place_id, place.sq_dist FROM (
SELECT DISTINCT(latlon_id) AS lat_lon_id, place_id, (@sq_dist:=(pow(x-@px, 2) + pow(y-@py, 2) + pow(z-@pz, 2))) AS sq_dist
FROM places_table WHERE id IN (
4030764,4030734,4030752,3948666,4030743,4030751,4030742,4030740,4030757,4030733,4030763,4030748,4030741,
4030735,4030744,4030753,4030737,4030736,4030731,8030076,4030739,6930873,4030727,4030758,4030726,8261466,
4030801,4030756,4030730,4030759,7840188,7911304,4030762,4030728,4030729,6531602,4030755,4030754,4030760,
4030749,4030750,4030761,8224616,4030738,4030732,4030746,4030747,4030745,4030871,4030872,4030790,4030787,
3948662,4030797,4030791,4030775,4030794,4030772,4030796,4030798,3948648,4030792,4030789,4030773,4030799,
3948661,3948651,4030788,4030778,3948657,4030800,4030795,4030793,4020117,4020363,3948663,4030777,3948658,
3948650,4030776,4020292,4020210,3948649,4030717,3969465,3969459,4030779,4030704,4030694,4030713,8529197,
4030873,3733656,3948664,4030786,4030781,4030783
) ORDER BY sq_dist LIMIT 1
) AS place, (
SELECT 0 AS index, 1636407.74908 AS x, -2220902.79092 AS y, -5744766.34094 AS z
UNION SELECT 1, 1674317.79921, -2157598.66673, -5757951.69661
UNION SELECT 2, 1652089.75753, -2193845.00579, -5750671.55762
UNION SELECT 3, 1621803.74283, -2184916.54092, -5762679.25265
UNION SELECT 4, 1615277.72619, -2200110.86847, -5758729.88373
UNION SELECT 5, 1652642.77785, -2208303.65375, -5744975.77555
UNION SELECT 6, 1618985.40684, -2190362.00049, -5761404.37734
UNION SELECT 7, 1621151.08717, -2208242.04656, -5753965.24636
UNION SELECT 8, 1663760.68219, -2166853.74959, -5757536.37073
UNION SELECT 9, 1639392.0856, -2136418.33191, -5775871.37502
) AS point ORDER BY point.index;
的价值由于某种原因无法计算，因此对它进行排序不会起作用。它给出了以下结果 -
@sq_dist我一直在用力撞击这一整夜，我尝试使用+-------+---------------+----------------+----------------+------------+------------+---------+
| index | point_x | point_y | point_z | lat_lon_id | place_id | sq_dist |
+-------+---------------+----------------+----------------+------------+------------+---------+
| 0 | 1636407.74908 | -2220902.79092 | -5744766.34094 | 433534 | 8529197 | NULL |
| 1 | 1674317.79921 | -2157598.66673 | -5757951.69661 | 433534 | 8529197 | NULL |
| 2 | 1652089.75753 | -2193845.00579 | -5750671.55762 | 433534 | 8529197 | NULL |
| 3 | 1621803.74283 | -2184916.54092 | -5762679.25265 | 433534 | 8529197 | NULL |
| 4 | 1615277.72619 | -2200110.86847 | -5758729.88373 | 433534 | 8529197 | NULL |
| 5 | 1652642.77785 | -2208303.65375 | -5744975.77555 | 433534 | 8529197 | NULL |
| 6 | 1618985.40684 | -2190362.00049 | -5761404.37734 | 433534 | 8529197 | NULL |
| 7 | 1621151.08717 | -2208242.04656 | -5753965.24636 | 433534 | 8529197 | NULL |
| 8 | 1663760.68219 | -2166853.74959 | -5757536.37073 | 433534 | 8529197 | NULL |
| 9 | 1639392.08560 | -2136418.33191 | -5775871.37502 | 433534 | 8529197 | NULL |
+-------+---------------+----------------+----------------+------------+------------+---------+
10 rows in set (0.10 sec)
，但我一直很难理解它的神秘方式。 应该使用GROUP BY只获得每个点的最小距离，但我似乎无法弄明白。
其他人可以想到一种让这项工作的方法吗？
任何帮助都将受到高度赞赏。
谢谢，
答案 0 :(得分：0)
所以，我使用GROUP BY自己想出来了。这是修改后的查询 -
SELECT t.latlon_id, t.place_id, t.latitude, t.longitude, t.indx, (@dist:=(pow(t.sq_dist, 0.5)/1000)) as dist from (
SELECT place.latlon_id, place.place_id, place.latitude, place.longitude, place.x, place.y, place.z, point.indx, point.xx, point.yy, point.zz,
(@sq_dist:=(pow(point.xx-place.x, 2) + pow(point.yy-place.y, 2) + pow(point.zz-place.z, 2))) AS sq_dist
FROM (
SELECT 0 AS indx, 1636407.74908 AS xx, -2220902.79092 AS yy, -5744766.34094 AS zz
UNION SELECT 1, 1674317.79921, -2157598.66673, -5757951.69661
UNION SELECT 2, 1652089.75753, -2193845.00579, -5750671.55762
UNION SELECT 3, 1621803.74283, -2184916.54092, -5762679.25265
UNION SELECT 4, 1615277.72619, -2200110.86847, -5758729.88373
UNION SELECT 5, 1652642.77785, -2208303.65375, -5744975.77555
UNION SELECT 6, 1618985.40684, -2190362.00049, -5761404.37734
UNION SELECT 7, 1621151.08717, -2208242.04656, -5753965.24636
UNION SELECT 8, 1663760.68219, -2166853.74959, -5757536.37073
UNION SELECT 9, 1639392.0856, -2136418.33191, -5775871.37502
) as point
JOIN geonames_geonamelookup place
WHERE place.geoname_id IN (
4030764,4030734,4030752,3948666,4030743,4030751,4030742,4030740,4030757,4030733,4030763,4030748,4030741,
4030735,4030744,4030753,4030737,4030736,4030731,8030076,4030739,6930873,4030727,4030758,4030726,8261466,
4030801,4030756,4030730,4030759,7840188,7911304,4030762,4030728,4030729,6531602,4030755,4030754,4030760,
4030749,4030750,4030761,8224616,4030738,4030732,4030746,4030747,4030745,4030871,4030872,4030790,4030787,
3948662,4030797,4030791,4030775,4030794,4030772,4030796,4030798,3948648,4030792,4030789,4030773,4030799,
3948661,3948651,4030788,4030778,3948657,4030800,4030795,4030793,4020117,4020363,3948663,4030777,3948658,
3948650,4030776,4020292,4020210,3948649,4030717,3969465,3969459,4030779,4030704,4030694,4030713,8529197,
4030873,3733656,3948664,4030786,4030781,4030783
) ORDER BY sq_dist
) as t GROUP BY t.indx;
MySQL GROUP BY根本不直观......
编辑： 请注意，以上查询速度非常慢且根本无法扩展，目前，我正在从python代码中执行的操作似乎运行良好且快速是，
count = len(xyz_s)
dist_phrase = ', (pow(x-%s, 2) + pow(y-%s, 2) + pow(z-%s, 2))' * count
query = """
select place_id, latlon_id, latitude, longitude {dist_phrase} from place_table where place_id in ({nbrs})
""".format(dist_phrase=dist_phrase, nbrs=nbrs) % tuple(a for xyz in xyz_s for a in xyz)
cursor.execute(query)
现在这给我的行数等于要查找的位数，以及包含每个点的平方距离的其他列，我进一步确定了python本身中每个点的最近位置。
希望这有助于某人。