我该如何处理在mysql中搜索记录的这种情况？答案

【问题标题】：How do i deal with this situation for searching records in mysql?我该如何处理在mysql中搜索记录的这种情况？
【发布时间】：2011-07-23 12:06:38
【问题描述】：

我正在开发一个 PHP/MYSQL 搜索模块，我必须根据许多不同的条件搜索表，我有大约 11 个表，并且我使用多个连接来创建一个 MySQL 查询并基于我打算的 WHERE 子句要搜索特定记录，这是我正在使用的 MYSQL 查询。

SELECT
prop.id,
prop.serial,
prop.title,
prop.rating,
prop.addDate,
prop.approve,
prop.status,
prop.user_id as userId,
user_det.email as email,
user_det.name as name,
prop.area_id as areaId,
area.name as areaName,
area.zipCode as zipCode,
area.city_id as cityId,
city.name as cityName,
city.state_id as stateId,
state.name as stateName,
state.country_id as countryId,
country.name as countryName,
prop.subCategory_id as subCategoryId,
subCat.name as subCategoryName,
subCat.category_id as categoryId,
cat.name as categoryName,
prop.transaction_id as transactionId,
trans.name as transactionName,
price.area as landArea,
price.price as priceSqFt,
price.total_price as totalPrice,
features.bedroom,
features.bathroom,
features.balcony,
features.furnished,
features.floorNum,
features.totalFloor
FROM properties prop 
LEFT JOIN user_details user_det ON (prop.user_id = user_det.user_id) 
LEFT JOIN areas area ON (prop.area_id = area.id) 
LEFT JOIN cities city ON (area.city_id = city.id) 
LEFT JOIN states state ON (city.state_id = state.id) 
LEFT JOIN countries country ON (state.country_id = country.id) 
LEFT JOIN subCategories subCat ON (prop.subCategory_id = subCat.id) 
LEFT JOIN categories cat ON (subCat.category_id = cat.id) 
LEFT JOIN transactions trans ON (prop.transaction_id = trans.id) 
LEFT JOIN prop_prices price ON (price.property_id = prop.id) 
LEFT JOIN prop_features features ON (features.property_id = prop.id)

虽然这里一切正常，但我有一个名为prop_amenities 的表，下面是该表的内容。

由于上表有多个property_id，如果我使用JOINS 查询它，那么根据我使用的JOIN 的类型，大多数情况下它将返回重复记录或省略其他记录的单个记录。所以我想这样处理。

使用表prop_amenities 只处理不返回结果的条件。例如，我正在搜索设施 ID 为 1、5、9、17 和 24 的属性，那么它应该检查 prop_amenities 表中是否存在所有记录，即在这种情况下为 1、5、9、17 和 24 .并返回所有上述选定列的相应记录。

我对使用 MySQL 处理这种情况一无所知。我该怎么做？

谢谢你..

【问题讨论】：

标签： mysql sql database

【解决方案1】：

您说“检查prop_amenities 表中的所有记录是否存在”，这就是这里的关键词。

SELECT ...
FROM properties AS prop
LEFT JOIN ...
WHERE EXISTS (SELECT 1 FROM prop_amenities AS pa WHERE pa.property_id = prop.property_id AND pa.amenity_id = 7);

【讨论】：