MySQL查詢與緯度經度和多個搜索加入

你好，我有MySQL查詢一樣，它的正常工作，但我想以檢查條件現在branches.visible = 1，則也顯示在查詢MySQL查詢與緯度經度和多個搜索加入

SELECT 
    company.name, 
    company.id, 
    company.mobile, 
    company.telephone, 
    IFNULL(fav.status, 'no') AS 'favorite',branchInfo.logo, 
    ((((acos(sin(($lat*pi()/180)) * sin((branches.lat*pi()/180))+cos(($lat*pi()/180)) * cos((branches.lat*pi()/180)) * cos((($lng- branches.lng)*pi()/180))))*180/pi())*60*1.1515*1.609344)) AS 'distance', 
    branchInfo.telephone, 
    branchInfo.id, 
    branchInfo.name, 
    branches.buffet, 
    branches.halal, 
    branches.kidsPlayArea, 
    branches.liveBand, 
    branches.outdoor, 
    branches.sheesha, 
    branches.valetParking, 
    branches.petsAllowed, 
    branches.wiFi, 
    branches.sportEvents, 
    branches.reservationRequired, 
    branches.wheelChairAccess, 
    branches.acceptCreditCard, 
    branches.brunch, 
    branchInfo.name, 
    branches.street, 
    branches.city, 
    branches.area, 
    branches.lat, 
    branches.lng, 
    branches.id, 
    branches.visible 
FROM branches 
LEFT JOIN users branchInfo 
    ON branchInfo.id=branches.branchId 
LEFT JOIN users company 
    ON company.id=branches.companyId 
LEFT JOIN favorite_branches fav 
    ON fav.branchId=branches.branchId AND fav.userId=$uId 
HAVING distance < 25

來源

2017-05-08 Muhammad irfan Mayo

你傷了我的眼睛（ – mcklayin

爲什麼我打破了你的眼睛兄弟 –

格式化你的代碼是可讀的東西，在發佈前考慮我已經編輯您的文章，以使代碼可讀。您是否嘗試添加where子句？ –

使用where子句，並添加您的條件有：

SELECT 
    company.name, 
    company.id, 
    company.mobile, 
    company.telephone, 
    IFNULL(fav.status, 'no') AS 'favorite',branchInfo.logo, 
    ((((acos(sin(($lat*pi()/180)) * sin((branches.lat*pi()/180))+cos(($lat*pi()/180)) * cos((branches.lat*pi()/180)) * cos((($lng- branches.lng)*pi()/180))))*180/pi())*60*1.1515*1.609344)) AS 'distance', 
    branchInfo.telephone, 
    branchInfo.id, 
    branchInfo.name, 
    branches.buffet, 
    branches.halal, 
    branches.kidsPlayArea, 
    branches.liveBand, 
    branches.outdoor, 
    branches.sheesha, 
    branches.valetParking, 
    branches.petsAllowed, 
    branches.wiFi, 
    branches.sportEvents, 
    branches.reservationRequired, 
    branches.wheelChairAccess, 
    branches.acceptCreditCard, 
    branches.brunch, 
    branchInfo.name, 
    branches.street, 
    branches.city, 
    branches.area, 
    branches.lat, 
    branches.lng, 
    branches.id, 
    branches.visible 
FROM branches 
LEFT JOIN users branchInfo 
    ON branchInfo.id=branches.branchId 
LEFT JOIN users company 
    ON company.id=branches.companyId 
LEFT JOIN favorite_branches fav 
    ON fav.branchId=branches.branchId AND fav.userId=$uId 
WHERE ((((acos(sin(($lat*pi()/180)) * sin((branches.lat*pi()/180))+cos(($lat*pi()/180)) * cos((branches.lat*pi()/180)) * cos((($lng- branches.lng)*pi()/180))))*180/pi())*60*1.1515*1.609344)) < 25 
AND branches.visible = 1

來源

2017-05-08 11:25:02

好抓@Strawberry。我已經更新了答案。請考慮改變你的倒票。 –

MySQL查詢與緯度經度和多個搜索加入

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