两个 table 一个包含记录和第二个包含位置我想显示第一个 table 记录而不加入

Question

我正在处理两个 table 包含记录；一个是管理员 table，其中包含所有详细信息，第二个是商店的位置，首先我想检查纬度和经度并从 shop_location 获得结果它工作正常，然后，我想检查正在营业的商店的状态并显示结果

这是我的 MYSQL QUERY，效果很好

SELECT DISTINCT s_number,radious, ( 3959 * ACOS( COS( RADIANS( 
   latitude ) ) * COS( RADIANS( "33.5967558" ) ) * COS( RADIANS( 
    "73.3869814" ) - RADIANS( longitude ) ) + SIN( RADIANS( latitude 
  ) ) * SIN( RADIANS("33.5967558" ) ) ) ) AS distance FROM 
   Shop_Location HAVING distance <= radius/1000 ORDER BY distance 
     ASC

这是我的第二个问题

SELECT * from Admins WHERE shopstatus='ON'

两个 table 中的电话号码相同但不是唯一的我想从 shop_location 查询中获取结果，然后检查商店状态 (ON) 显示结果。

我想用Query解决

Answer 1

假设两个表都链接到“s_number”（店号），您可以尝试以下SQL

SELECT DISTINCT Shop_Location.s_number,radious, ( 3959 * ACOS( COS( RADIANS( 
   latitude ) ) * COS( RADIANS( "33.5967558" ) ) * COS( RADIANS( 
    "73.3869814" ) - RADIANS( longitude ) ) + SIN( RADIANS( latitude 
  ) ) * SIN( RADIANS("33.5967558" ) ) ) ) AS distance FROM 
   Shop_Location, admins where
Shop_Location.s_number=admins.s_number and admins.shopstatus='ON'
 HAVING distance <= radius/1000 ORDER BY distance 
     ASC

Answer 2

可能是这样的：

SELECT DISTINCT
    Shop_location.s_number,
    Shop_location.radious,
    (
        3959 * ACOS(
            COS(RADIANS(latitude)) * COS(RADIANS("33.5967558")) * COS(
                RADIANS("73.3869814") - RADIANS(Shop_location.longitude)
            ) + SIN(
                RADIANS(Shop_location.latitude)
            ) * SIN(RADIANS("33.5967558"))
        )
    ) AS DISTANCE,
    Admins.shopstatus
FROM
    Shop_Location
LEFT JOIN
    Admins
ON
    Admins.s_number = Shop_location.s_number
HAVING
    DISTANCE <= radius / 1000
ORDER BY
    DISTANCE ASC

Answer 3

这是正确的答案

SELECT DISTINCT 
Admins.email,Admins.address,Admins.adminname,
Admins.diliveryfee,Admins.endingtime,Admins.minorder
,Admins.name,Admins.password,Admins.shopimage,
Admins.shopmessage,Admins.phonenumber,
 Admins.shopname,Admins.shopstatus FROM (
 SELECT Admins.email,Admins.address,Admins.adminname,Admins.diliveryfee,Admins.endingtime,Admins.minorder,Admins.name,Admins.password,Admins.shopimage,Admins.shopmessage,Admins.phonenumber,Admins.shopname,Admins.shopstatus,Shop_Location.radious AS shopLocationRadius, ( 6371 * ACOS( COS( 
  RADIANS( latitude ) ) * COS( RADIANS( "33.5967558" ) ) * 
  COS( RADIANS( "73.3869814" ) - RADIANS( longitude ) ) + 
    SIN( RADIANS( latitude ) ) * SIN( 
     RADIANS("33.5967558" ) 
      ) ) ) AS distance FROM Shop_Location INNER JOIN 
    Admins ON 
    (Admins.phonenumber = Shop_Location.s_number) WHERE 
    Admins.shopstatus='ON'  HAVING distance <= radious/1000 
   ORDER BY distance ASC)
     AS Admins

两个 table 一个包含记录和第二个包含位置我想显示第一个 table 记录而不加入

Two tables one contain records and second conatain location i want to show first table record with out join

php

mysql

jquery

mysql-error-1064