查詢排行及所有(表名.*)

1.

set @rownum=0;SELECT @rownum:=@rownum+1 AS top, customer.* FROM customer

?

2.

SELECT @rownum:=@rownum+1 AS top, customer.* FROM (SELECT @rownum:=0) r, customer ORDER BY customer.this_month_cost DESC

1和2的結(jié)果：

查詢排行及id(表名.id)

SELECT @rownum:=@rownum+1 AS top, customer.id FROM (SELECT @rownum:=0) r, customer ORDER BY customer.this_month_cost DESC

結(jié)果：

排序后更新原表的排行：

UPDATE `customer` c,(SELECT@rownum :=@rownum + 1 AS rownum,customer.idFROM(SELECT @rownum := 0) r,customerORDER BYcustomer.this_month_cost DESC ) a SET c.top = a.rownum WHEREc.id = a.id

?

轉(zhuǎn)載于:https://www.cnblogs.com/Alwaysbecoding/p/6767695.html

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