Question

下面我有一个查询，可以查找距离表$dist = 100;

一定距离内的人（'persons'内）

$origLat = '8.574596';
$origLon = '-73.602050';
$dist = 100; // This is the maximum distance (in miles) away from $origLat, $origLon in which to search
$query = "SELECT id_the_person, Name, Lat, Lng, 3959 * 2 * 
          ASIN(SQRT( POWER(SIN(($origLat - Lat)*pi()/180/2),2)
          +COS($origLat*pi()/180 )*COS(Lat*pi()/180)
          *POWER(SIN(($origLon-Lng)*pi()/180/2),2))) 
          as distance FROM persons WHERE 
          Lng between ($origLon-$dist/cos(radians($origLat))*69) 
          and ($origLon+$dist/cos(radians($origLat))*69) 
          and Lat between ($origLat-($dist/69)) 
          and ($origLat+($dist/69))
          having distance < $dist ORDER BY distance limit 100";

在此查询中，我根据他们在表'svar'

中回答问题时的匹配方式找到了人

SELECT *, round(count(*)/(SELECT count(*) FROM svar where id_the_person = ?)*100) as pecentage
FROM svar as l
where l.svaret=(
    SELECT m.svaret
    FROM svar as m
    where m.id_the_person = ? and l.id_frag=m.id_frag)
group by (l.id_the_person)
order by pecentage");

两个表中的变量id_the_person相同。我想要的是根据第一个查询匹配短按百分比但只有距离内的人的人，即第二个查询仅限于$dist=100内的人。

使用公共变量从两个表中连接两个sql查询

0 个答案: