mysql 根据某一列进行区间统计
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mysql中可以利用elt,interval函数来实现此类需求
mysql> select * from k1; +------+------+ | id | yb | +------+------+ | 1 | 100 | | 2 | 11 | | 3 | 5 | | 4 | 501 | | 5 | 1501 | | 6 | 1 | +------+------+
现在要进行统计,小于100的,100~500的,500~1000的,1000以上的,这各个区间的id数
利用interval划出4个区间,再利用elt函数将4个区间分别返回一个列名:
mysql> select elt(interval(d.yb,0, 100, 500, 1000), ‘less100‘, ‘100to500‘, ‘500to1000‘, ‘more1000‘) as yb_level, count(d.id) as cnt from k1 d group by elt(interval(d.yb, 0, 100, 500, 1000), ‘less100‘, ‘100to500‘, ‘500to1000‘, ‘more1000K‘); +-----------+-----+ | yb_level | cnt | +-----------+-----+ | 100to500 | 1 | | 500to1000 | 1 | | less100 | 3 | | more1000 | 1 | +-----------+-----+ 4 rows in set (0.00 sec)
如果需要按从小到大排序的话 可以在列名定义时稍加一个首字符 对各档区间进行排序
mysql>select elt(interval(d.yb,0, 100, 500, 1000), ‘1/less100‘, ‘2/100to500‘, ‘3/500to1000‘, ‘4/more1000‘) as yb_level, count(d.id) as cnt from k1 d group by elt(interval(d.yb, 0, 100, 500, 1000), ‘1/less100‘, ‘2/100to500‘, ‘3/500to1000‘, ‘4/more1000K‘); +-------------+-----+ | yb_level | cnt | +-------------+-----+ | 1/less100 | 3 | | 2/100to500 | 1 | | 3/500to1000 | 1 | | 4/more1000 | 1 | +-------------+-----+ 4 rows in set (0.00 sec)
附elt函数格式:
ELT(N,str1,str2,str3,...)
如果N= 1,返回str1,如果N= 2,返回str2,等等。如果N小于1或大于参数个数,返回NULL。ELT()是FIELD()反运算。
mysql> select ELT(1, ‘ej‘, ‘Heja‘, ‘hej‘, ‘foo‘); -> ‘ej‘ mysql> select ELT(4, ‘ej‘, ‘Heja‘, ‘hej‘, ‘foo‘); -> ‘foo‘
interval函数格式:
INTERVAL() Return the index of the argument that is less than the first argument(小于后面的某个参数,就返回这个参数的前一个位置数字)
INTERVAL(N,N1,N2,N3,...)
Returns 0 if N < N1, 1 if N < N2 and so on or -1 if N is NULL. All arguments are treated as integers. It is required that N1 < N2 < N3 < ... < Nn for this function to work correctly. This is because a binary search is used (very fast).
mysql> SELECT INTERVAL(23, 1, 15, 17, 30, 44, 200); (23小于30,30的位置是4,于是返回3) -> 3 mysql> SELECT INTERVAL(10, 1, 10, 100, 1000); -> 2 mysql> SELECT INTERVAL(22, 23, 30, 44, 200); -> 0
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