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117. To get around this issue, wrap your select statement in a CTE, and then you can query against the CTE and use the windowed function's results in the where clause. WITH MyCte AS. (. select employee_id, RowNum = row_number() OVER ( order by employee_id ) from V_EMPLOYEE. ORDER BY Employee_ID.
3. The & operator in Java is the bitwise-and operator. Basically, (number & 1) performs a bitwise-and between number and 1. The result is either 0 or 1, depending on whether it's even or odd. Then the result is compared with 0 to determine if it's even. Here's a page describing bitwise operations.
Task: generate random number between 1 and 6. Math.random() returns floating point number between 0 and 1 (like 0.344717274374 or 0.99341293123 for example), which we will use as a percentage, so Math.floor(Math.random() * 6) + 1 returns some percentage of 6 (max: 5, min: 0) and adds 1.
SQL Row_Number () function is to sort and assign an order number to data rows in related record set. So it is used to number rows, for example to identify the top 10 rows which have the highest order amount or identify the order of each customer which is the highest amount, etc. If you want to sort the dataset and number each row by seperating ...
While list(map(int, str(x))) is the Pythonic approach, you can formulate logic to derive digits without any type conversion: from math import log10. def digitize(x): n = int(log10(x)) for i in range(n, -1, -1): factor = 10**i. k = x // factor. yield k. x -= k * factor.
Java 7+. In Java 1.7 or later, the standard way to do this (generate a basic non-cryptographically secure random integer in the range [min, max]) is as follows: import java.util.concurrent.ThreadLocalRandom; // nextInt is normally exclusive of the top value, // so add 1 to make it inclusive. int randomNum = ThreadLocalRandom.current().nextInt ...
You can easily give a counter example: create table ex(n number); insert into ex(n) values(1.5); select * from ex; You get back 1.5. With scale=0 you would get back 1. This is also documented, e.g. with 11.2 Oracle documents that just NUMBER (without precision and without scale) results in 'maximum range and precision'. – maxschlepzig.
I've got a solution that counts the bits in O(Number of 1's) time: bitcount(n): count = 0. while n > 0: count = count + 1. n = n & (n-1) return count. In worst case (when the number is 2^n - 1, all 1's in binary) it will check every bit. Edit: Just found a very nice constant-time, constant memory algorithm for bitcount.
Random rand = new Random(); // nextInt as provided by Random is exclusive of the top value so you need to add 1 int randomNum = rand.nextInt((max - min) + 1) + min; See the relevant JavaDoc . As explained by Aurund, Random objects created within a short time of each other will tend to produce similar output, so it would be a good idea to keep ...
while. That’s because ToNumber (null) is specified to return 0. parseInt, however expects strings, so null and undefined are coerced to the strings "null" or "undefined", respectively, first, which don’t start with digits. Number(undefined) is NaN, not 0, which is a special step of the Number(…) algorithm.