Could someone explain the difference between % in SQL?
I understand that % is a wildcard that allows you to query results with LIKE results, i.e. a% for words starting with a, but I am confused why the wildcard can be used as % 2 = 0 to query for even numbers?
I saw an explanation that said % can be used as divide but I thought / was divide.
a % 2 = 0 here % as Modulus arithmetic operator.
Syntax: dividend % divisor
Sample: SELECT 15 % 2 AS Remainder it will return the result as 1
Demo on db<>fiddle
When used outside of a string, the percentage symbol % is the modulus operator, i.e. an operator which returns the remainder following division of the number preceding the operator by that following it.
Therefore, in your example, the expression % 2 = 0 will be validated if the number preceding the percentage symbol is even, e.g. 12 % 2 = 0 will return True.
Whereas, when used in the pattern argument of a like expression, the percentage symbol represents a wildcard operator matching any sequence of characters (or no characters at all).
Let's understand with an example:
I have created an Table name - 'c', which contain 2 attribute 'name' and 'num'.
when num%10 is calculated e.g. 55%10 -> gives 5.
If result is either 2 or 7 then it will not print that row
Elseif result (num%10) is NOT 2 or 7 then in this case it will print the row.
Now:
Select *from c where num%10 NOT In(2,7);
Check out Screenshot here :enter image description here
I have a bit of math question for my sql code. I want to be able to work out the following in my code using a mathematical formula, if anyone knows how I would love to know.
Any number 1->99 : 10
Any number 100->999 : 100
Any number 1000->9999 : 1000
...
Is there anyway to work out the 10 multiplier just from the value? I feel like this should be an easy formula but I cant seem to get it.
Thanks
How about this?
SELECT POWER(10, CONVERT(INT, LOG10(#Input)))
It takes the log base 10 of the input value (which returns the value of the exponent to which you would have to raise 10 to in order to get the input value), then it lops off the decimal portion leaving only the whole number, and then raises 10 to that power.
You just need logs and their opposite (power)...
power(10, floor(log10(x)))
As follows...
log10(99) = 1.9956351946
floor(1.9956351946) = 1
power(10, 1) = 10
This does, however, assume that your example is wrong and that 1 -> 9 should "round" to 1...
log10(9) = 0.95424250943
floor(0.95424250943) = 0
power(10, 0) = 1
I need to divide one int into 2 other int's. the first int is not constant so one problem would be, what to do with odd numbers because I only want whole numbers. For example, if int = 5, then int(2) will = 2 and int(3) will = 3. Any help will greatly be appreciated.
Supposing you want to express x = a + b, where a and b are as close to x/2 as possible:
a = ceiling(x / 2.0);
b = floor(x / 2.0);
That's pseudo code, you have to find out the actual functions for floor and ceiling from your library. Make sure the division is performed as floating point numbers.
As pure integers:
a = x / 2 + (x % 2 == 0 ? 0 : 1);
b = x / 2
(This may be a bit fishy for negative numbers, because it'll depend on the behaviour of division and modulo for negative numbers.)
You can try ceil and floor functions from math to produce results like 2 and 3 for odd inputs;
int(2)=ceil(int/2); //will produce 3 for input 5
int(3)=floor(int/2); //will produce 2 for input 5
Well my answer is not in Objective-C but i guess you could translate this easily.
My idea is:
part1 = source_number div 2
part2 = source_number div 2 + (source_number mod 2)
This way the second number will be bigger if the starting number is an odd number.
I'm trying to make a generic equation which converts a value. Here are some examples.
9,873,912 -> 9,900,000
125,930 -> 126,000
2,345 -> 2,400
280 -> 300
28 -> 30
In general, x -> n
Basically, I'm making a graph and I want to make values look nicer. If it's a 6 digit number or higher, there should be at least 3 zeros. If it's a 4 digit number or less, there should be at least 2 digit numbers, except if it's a 2 digit number, 1 zero is fine.
(Ignore the commas. They are just there to help read the examples). Anyways, I want to convert a value x to this new value n. What is an equation g(x) which spits out n?
It is for an objective-c program (iPhone app).
Divide, truncate and multiply.
10**x * int(n / 10**(x-d))
What is "x"? In your examples it's about int(log10(n))-1.
What is "d"? That's the number of significant digits. 2 or 3.
Ahhh rounding is a bit awkward in programming in general. What I would suggest is dividing by the power of ten, int cast and multiplying back. Not remarkably efficient but it will work. There may be a library that can do this in Objective-C but that I do not know.
if ( x is > 99999 ) {
x = ((int)x / 1000) * 1000;
}
else if ( x > 999 ) {
x = ((int) x / 100) * 100;
}
else if ( x > 9 ) {
x = ((int) x / 10) * 10;
}
Use standard C functions like round() or roundf()... try man round at a command line, there are several different options depending on the data type. You'll probably want to scale the values first by dividing by an appropriate number and then multiplying the result by the same number, something like:
int roundedValue = round(someNumber/scalingFactor) * scalingFactor;
I know the modulus (%) operator calculates the remainder of a division. How can I identify a situation where I would need to use the modulus operator?
I know I can use the modulus operator to see whether a number is even or odd and prime or composite, but that's about it. I don't often think in terms of remainders. I'm sure the modulus operator is useful, and I would like to learn to take advantage of it.
I just have problems identifying where the modulus operator is applicable. In various programming situations, it is difficult for me to see a problem and realize "Hey! The remainder of division would work here!".
Imagine that you have an elapsed time in seconds and you want to convert this to hours, minutes, and seconds:
h = s / 3600;
m = (s / 60) % 60;
s = s % 60;
0 % 3 = 0;
1 % 3 = 1;
2 % 3 = 2;
3 % 3 = 0;
Did you see what it did? At the last step it went back to zero. This could be used in situations like:
To check if N is divisible by M (for example, odd or even)
or
N is a multiple of M.
To put a cap of a particular value. In this case 3.
To get the last M digits of a number -> N % (10^M).
I use it for progress bars and the like that mark progress through a big loop. The progress is only reported every nth time through the loop, or when count%n == 0.
I've used it when restricting a number to a certain multiple:
temp = x - (x % 10); //Restrict x to being a multiple of 10
Wrapping values (like a clock).
Provide finite fields to symmetric key algorithms.
Bitwise operations.
And so on.
One use case I saw recently was when you need to reverse a number. So that 123456 becomes 654321 for example.
int number = 123456;
int reversed = 0;
while ( number > 0 ) {
# The modulus here retrieves the last digit in the specified number
# In the first iteration of this loop it's going to be 6, then 5, ...
# We are multiplying reversed by 10 first, to move the number one decimal place to the left.
# For example, if we are at the second iteration of this loop,
# reversed gonna be 6, so 6 * 10 + 12345 % 10 => 60 + 5
reversed = reversed * 10 + number % 10;
number = number / 10;
}
Example. You have message of X bytes, but in your protocol maximum size is Y and Y < X. Try to write small app that splits message into packets and you will run into mod :)
There are many instances where it is useful.
If you need to restrict a number to be within a certain range you can use mod. For example, to generate a random number between 0 and 99 you might say:
num = MyRandFunction() % 100;
Any time you have division and want to express the remainder other than in decimal, the mod operator is appropriate. Things that come to mind are generally when you want to do something human-readable with the remainder. Listing how many items you could put into buckets and saying "5 left over" is good.
Also, if you're ever in a situation where you may be accruing rounding errors, modulo division is good. If you're dividing by 3 quite often, for example, you don't want to be passing .33333 around as the remainder. Passing the remainder and divisor (i.e. the fraction) is appropriate.
As #jweyrich says, wrapping values. I've found mod very handy when I have a finite list and I want to iterate over it in a loop - like a fixed list of colors for some UI elements, like chart series, where I want all the series to be different, to the extent possible, but when I've run out of colors, just to start over at the beginning. This can also be used with, say, patterns, so that the second time red comes around, it's dashed; the third time, dotted, etc. - but mod is just used to get red, green, blue, red, green, blue, forever.
Calculation of prime numbers
The modulo can be useful to convert and split total minutes to "hours and minutes":
hours = minutes / 60
minutes_left = minutes % 60
In the hours bit we need to strip the decimal portion and that will depend on the language you are using.
We can then rearrange the output accordingly.
Converting linear data structure to matrix structure:
where a is index of linear data, and b is number of items per row:
row = a/b
column = a mod b
Note above is simplified logic: a must be offset -1 before dividing & the result must be normalized +1.
Example: (3 rows of 4)
1 2 3 4
5 6 7 8
9 10 11 12
(7 - 1)/4 + 1 = 2
7 is in row 2
(7 - 1) mod 4 + 1 = 3
7 is in column 3
Another common use of modulus: hashing a number by place. Suppose you wanted to store year & month in a six digit number 195810. month = 195810 mod 100 all digits 3rd from right are divisible by 100 so the remainder is the 2 rightmost digits in this case the month is 10. To extract the year 195810 / 100 yields 1958.
Modulus is also very useful if for some crazy reason you need to do integer division and get a decimal out, and you can't convert the integer into a number that supports decimal division, or if you need to return a fraction instead of a decimal.
I'll be using % as the modulus operator
For example
2/4 = 0
where doing this
2/4 = 0 and 2 % 4 = 2
So you can be really crazy and let's say that you want to allow the user to input a numerator and a divisor, and then show them the result as a whole number, and then a fractional number.
whole Number = numerator/divisor
fractionNumerator = numerator % divisor
fractionDenominator = divisor
Another case where modulus division is useful is if you are increasing or decreasing a number and you want to contain the number to a certain range of number, but when you get to the top or bottom you don't want to just stop. You want to loop up to the bottom or top of the list respectively.
Imagine a function where you are looping through an array.
Function increase Or Decrease(variable As Integer) As Void
n = (n + variable) % (listString.maxIndex + 1)
Print listString[n]
End Function
The reason that it is n = (n + variable) % (listString.maxIndex + 1) is to allow for the max index to be accounted.
Those are just a few of the things that I have had to use modulus for in my programming of not just desktop applications, but in robotics and simulation environments.
Computing the greatest common divisor
Determining if a number is a palindrome
Determining if a number consists of only ...
Determining how many ... a number consists of...
My favorite use is for iteration.
Say you have a counter you are incrementing and want to then grab from a known list a corresponding items, but you only have n items to choose from and you want to repeat a cycle.
var indexFromB = (counter-1)%n+1;
Results (counter=indexFromB) given n=3:
`1=1`
`2=2`
`3=3`
`4=1`
`5=2`
`6=3`
...
Best use of modulus operator I have seen so for is to check if the Array we have is a rotated version of original array.
A = [1,2,3,4,5,6]
B = [5,6,1,2,3,4]
Now how to check if B is rotated version of A ?
Step 1: If A's length is not same as B's length then for sure its not a rotated version.
Step 2: Check the index of first element of A in B. Here first element of A is 1. And its index in B is 2(assuming your programming language has zero based index).
lets store that index in variable "Key"
Step 3: Now how to check that if B is rotated version of A how ??
This is where modulus function rocks :
for (int i = 0; i< A.length; i++)
{
// here modulus function would check the proper order. Key here is 2 which we recieved from Step 2
int j = [Key+i]%A.length;
if (A[i] != B[j])
{
return false;
}
}
return true;
It's an easy way to tell if a number is even or odd. Just do # mod 2, if it is 0 it is even, 1 it is odd.
Often, in a loop, you want to do something every k'th iteration, where k is 0 < k < n, assuming 0 is the start index and n is the length of the loop.
So, you'd do something like:
int k = 5;
int n = 50;
for(int i = 0;i < n;++i)
{
if(i % k == 0) // true at 0, 5, 10, 15..
{
// do something
}
}
Or, you want to keep something whitin a certain bound. Remember, when you take an arbitrary number mod something, it must produce a value between 0 and that number - 1.