根節點至少2顆子樹(若存在子樹的情況下);
非根節點至少擁有m/2顆子樹,其範圍為m/2 <= childnum(x) <= m;
所有葉子節點都在同一層,且為null;
有k顆子樹的節點,其關鍵字數為k-1,ceil(m/2)-1 <= keynum(x) <= m-1;
m/2+(m/2)^2+...+(m/2)^(h-1) = n
\frac = n
(m/2)^(h-1) = n+1
h = (log_(m/2) (n+1))+1
m+(m)^2+...+(m)^(h-1) = n
\frac = n
(m)^(h-1) = n+1
h = (log_(m) (n+1))+1
根節點至少2顆子樹(若存在子樹的情況下);
有n顆子樹的節點有n個關鍵字;
所有內節點僅存放索引,資料全部儲存在葉子節點上。
紅黑樹高度為35;
show index from pre_sales_rfq
其中id為主鍵,(id, name,create_date)聯合索引
explain select * from index_test where id = 1
and name = 'jackshawn'
and create_date = '2017-09-21'
explain select * from pre_sales_rfq where project_id = 1
and item_id = 1
explain select * from pre_sales_project_rfq where item_id = 1
and project_id = 1
explain select * from pre_sales_project_rfq where project_id = 1
(id, name,create_date)聯合索引
explain select * from index_test where id = 1
and create_date = '2017-09-21'
explain select * from pre_sales_project_rfq where item_id = 1
explain select * from index_test where id = 1
and name ='%jack'
explain select * from index_test where id = 1
and name ='jack%'
explain select * from index_test where id > 1
and name ='jackshawn'
explain select * from index_test where id in(1,4) and name ='jackshawn'
explain select * from index_test where id between 1
and4
and name ='jackshawn'
explain select * from index_test where id -1 = 1
and name ='jackshawn'
## 乙個欄位不重複的總量與資料總量的比值,越**擇性越好。
select
count(distinct(key))/count(*) as selectivity from
table;
2.查詢效率高。
## 舉例如下擷取last_name前4個字元
alter
table employees.employees
add index `first_name_last_name4` (first_name, last_name(4));
select
count(distinct(concat(first_name, left(last_name, 4))))/count(*) as selectivity from employees.employees;
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