Skip to main content

How to use Connect by oracle

first create table

create table node(
em varchar(100),
ma varchar(100)

insert into node (em, ma) values(upper('a'), null)
insert into node (em, ma) values(upper('b'), upper('a'))
insert into node (em, ma) values(upper('c'), upper('a'))
insert into node (em, ma) values(upper('d'), upper('c'))
insert into node (em, ma) values(upper('e'), upper('c'))


select * from node

select level, lpad(' ', 4 * (level-1)) || em 사원, ma 관리자, connect_by_isleaf isleaf
from node
start with ma is null
connect by prior em = ma

select connect_by_root em 루트사원, sys_connect_by_path(em, '/') 경로, em, ma
from node
start with ma is null
connect by prior em = ma


Popular posts from this blog

codefights smooth sailing ( CommonCharacterCount) s1, String s2) { int sum = 0; char[] as= s1.toCharArray(); char[] bs= s2.toCharArray(); int[] ias = newint[126]; int[] ibs = newint[126]; for (int i = 0; i < as.length; i++) { ias[(int)as[i]]++; } for (int i = 0; i < bs.length; i++) { ibs[(int)bs[i]]++; } for (int i = 0; i < ibs.length; i++) { sum += Math.min(ias[i], ibs[i]); } return sum; }

Given two strings, find the number of common characters between them. Example For s1 = "aabcc" and s2 = "adcaa", the output should be
commonCharacterCount(s1, s2) = 3. Strings have 3 common characters - 2 "a"s and 1 "c". Input/Output [time limit] 3000ms (java)[input] string s1 A string consisting of lowercase latin letters a-z. Guaranteed constraints:
1 ≤ s1.length ≤ 15. [input] string s2 A string consisting of lowercase latin letters a-z. Guaranteed constr…

Bucket Sort in python

I make buckets as many as size of arr
and put data.

arr = [0.897, 0.565, 0.656, 0.1234, 0.665, 0.3434] def bucketSort(arr, size): buckets = [[] for i in range(size)] # put arr in bucket for i in range(len(arr)): num = size*arr[i] buckets[int(num)].append(arr[i]) output = [] # use insertion sort for i in range(len(buckets)): insertionSort(buckets[i]) # concat all data for i in range(len(buckets)): while len(buckets[i]) > 0: output.append(buckets[i].pop(0)) return output def swap(arr, i, j): temp = arr[i] arr[i] = arr[j] arr[j] = temp def insertionSort(arr): for i in range(1, len(arr)): index= i print("index : " + str(i)) while index!=0: if arr[index] < arr[index-1]: temp = arr[index] arr[index]= arr[index-1] arr[index-1] = temp index = index-1print(arr) else …