我有两张table；songs有48959281行，popular_songs有5721117行。这是两个表的结构:CREATETABLEsongs(songIdBIGINT(20)NOTNULL,songNameVARCHAR(1000)NOTNULL,songDurationBIGINT(20)NOTNULL,songPreviewVARCHAR(1000)NOTNULL,PRIMARYKEY(songId),INDEX(songDuration))ENGINE=InnoDBDEFAULTCHARSET=utf8DEFAULTCOLLATEutf8_general_ci;CREAT
我在MySQL5.1.38中有两个表。products+----+------------+-------+------------+|id|name|price|department|+----+------------+-------+------------+|1|FireTruck|15.00|Toys||2|Bike|75.00|Toys||3|T-Shirt|18.00|Clothes||4|Skirt|18.00|Clothes||5|Pants|22.00|Clothes|+----+------------+-------+------------+ratings+-
我在MySQL5.1.38中有两个表。products+----+------------+-------+------------+|id|name|price|department|+----+------------+-------+------------+|1|FireTruck|15.00|Toys||2|Bike|75.00|Toys||3|T-Shirt|18.00|Clothes||4|Skirt|18.00|Clothes||5|Pants|22.00|Clothes|+----+------------+-------+------------+ratings+-
SELECTSUM(CASEWHENcumulative=1THENpercentELSE0END)FROMphppos_items_taxes;鉴于上述声明，它会执行以下操作:mysql>select*FROMphppos_items_taxes;+---------+-----------+---------+------------+|item_id|name|percent|cumulative|+---------+-----------+---------+------------+|1|Tax1|8.00|0||1|Tax2|10.00|1||3|SalesTax|8.0
SELECTSUM(CASEWHENcumulative=1THENpercentELSE0END)FROMphppos_items_taxes;鉴于上述声明，它会执行以下操作:mysql>select*FROMphppos_items_taxes;+---------+-----------+---------+------------+|item_id|name|percent|cumulative|+---------+-----------+---------+------------+|1|Tax1|8.00|0||1|Tax2|10.00|1||3|SalesTax|8.0
给定m2m关系:items-categories我有三个表:项目,类别和items_categories包含对两者的引用我想找到属于所有给定类别集的项目:FindItembelongingtoacategoryin[1,3,6]andbelongingtoacategoryin[7,8,4]andbelongingtoacategoryin[12,66,42]and...我可以想到两种方法在mySQL中实现这一点。选项A:内部联接:SELECTidfromitemsINNERJOINcategoryc1ON(item.id=c1.item_id)INNERJOINcategoryc2O
给定m2m关系:items-categories我有三个表:项目,类别和items_categories包含对两者的引用我想找到属于所有给定类别集的项目:FindItembelongingtoacategoryin[1,3,6]andbelongingtoacategoryin[7,8,4]andbelongingtoacategoryin[12,66,42]and...我可以想到两种方法在mySQL中实现这一点。选项A:内部联接:SELECTidfromitemsINNERJOINcategoryc1ON(item.id=c1.item_id)INNERJOINcategoryc2O
所以我得到了一组看起来像这样的结果:SELECTUser_ID,StartTime,EndTime,TIMEDIFF(EndTime,StartTime)ASTimeDiffFROMMyTable------------------------------------------------------------------|User_ID|StartTime|EndTime|TimeDiff|------------------------------------------------------------------|1|2010-11-0508:00:00|2010-11-
所以我得到了一组看起来像这样的结果:SELECTUser_ID,StartTime,EndTime,TIMEDIFF(EndTime,StartTime)ASTimeDiffFROMMyTable------------------------------------------------------------------|User_ID|StartTime|EndTime|TimeDiff|------------------------------------------------------------------|1|2010-11-0508:00:00|2010-11-
我在表格中有类别表格(猫名，金额)；如何得到amount每个cat_name按cat_name分组的总和 最佳答案 SELECTcat_name,SUM(amount)AStotal_amountFROMtableGROUPBYcat_name 关于MySQL按SUM分组，我们在StackOverflow上找到一个类似的问题： https://stackoverflow.com/questions/6105767/
