I am trying to generate a random value between 0.01 - 0.50 to enter into mysql. I have 2.7 million rows that I need to execute this on.
Here is my script:
UPDATE FBAInventory SET buyBox = ROUND( 0.01 + RAND( ) * 8,2 );
It is generating values such as 4.20, 3.89 etc. I only want it to span from 0.01 - 0.50 and not to exceed this.
Does anyone know how to do this?
Thanks!
How about...
round(rand() * 0.49 + 0.01, 2);
You can use the floor function to generate a range of random numbers.
FLOOR(RAND() * (<max> - <min> + 1)) + <min>
where the max and min are inclusive. So in your case you would want
FLOOR(RAND() * 1.49 ) + 0.01
Related
I'm using MySQL version: 5.7.22
I've got two columns Latitude and Longitude both in Degrees/Minutes/Seconds format that I want to convert to Decimal format ex: 48° 52.250' N to 48.93611111
I have the following script but I'm stuck at how to split the degrees minutes and seconds. I cannot hard-code the values as I've done here left(Latitude,2) since the degrees might have 3 decimals as well
SELECT Latitude,
left(Latitude, 2) +
(TRUNCATE((Latitude - TRUNCATE(Latitude)) * 100) / 60) +
(((Latitude * 100) - TRUNCATE(Latitude * 100)) * 100) / (60 * 60) AS DECIMAL_DEGREES
FROM small_ocean_data
The formula for the conversion is this: D + M/60 + S/3600 * -1 if direction in ['W', 'S'] else 1
Any help would be grateful!
Well if I use this formula of yours: D + M/60 + S/3600
Where I believe D is Degrees and M are Minutes and S are Seconds.
With this select:
select SUBSTRING_INDEX('48° 52.250', '°',1) +
(SUBSTRING_INDEX('48° 52.250',' ',1)/60) +
(SUBSTRING_INDEX('48° 52.250','.',1)/3600);
Or for your database:
select SUBSTRING_INDEX(Latitude, '°', 1) +
(SUBSTRING_INDEX(Latitude ,' ',1)/60) +
(SUBSTRING_INDEX(Latitude ,'.',1)/3600) "DECIMAL_DEGREES"
FROM small_ocean_data;
I get 48.81333333333333
48 + 0.8 + 0.013333333333333334
Here is the DEMO
I want to use SQL's RAND()-function to get the number 400, 450 or 500 and insert it in a field inside a tabe.
How could I do that?
I think you can follow code bellow
a = floor(rand() * 3);
value = 400 + a*50;
First genterate ints from set [0,1,2] then multiple to 50 to get step in 50 then add seed
select floor(rand() * 3) * 50 + 400
Assume a service is billed in the following manner:
The first 60 seconds is charged at $1.00
Subsequent charges are billed at $0.25 per 10 second
The following are example computations:
32 seconds = $1.00
59 seconds = $1.00
60 seconds = $1.00
61 seconds = $1.25
69 seconds = $1.25
70 seconds = $1.25
71 seconds = $1.50
Is it possible to do this kind of computation in MySQL alone?
EDIT 1:
Does something like this work:
SELECT `call_length`,
( 1.00 + ( Round(( `call_length` - 30 ) / 10) * .25 ) ) AS `cost`
FROM `service`
SqlFiddleDemo
CREATE TABLE sec(val INT);
INSERT INTO sec
VALUES (32), (59), (60), (61), (69), (70), (71);
SELECT
val,
1.0 + CASE
WHEN val <= 60.0 THEN 0
WHEN val MOD 10 = 0 THEN 0.25 *((val - 60) DIV 10)
ELSE 0.25 * (((val - 60) DIV 10) + 1)
END AS charge
FROM sec;
EDIT:
Without CASE:
SqlFiddleDemo2
SELECT
call_length,
1.0 + IF( call_length <= 60, 0, 0.25 * CEIL((call_length - 60)/10)) AS cost
FROM service;
This is not much of a MySQL problem, unless the setting in which you need to perform the calculation is somehow difficult(?).
UPDATE ... SET cost_cents = 100 + CEIL(GREATEST(0, duration - 60)/10) * 25;
As a SELECT to match your edit,
SELECT `call_length`,
100 + CEIL(GREATEST(0, `call_length` - 60)/10) * 25 AS `cost`
FROM `service`
Note that this returns cents. For dollars, divide the result by 100...
SELECT `call_length`,
(100 + CEIL(GREATEST(0, `call_length` - 60)/10) * 25) / 100 AS `cost`
FROM `service`
I want to convert feet and inches to centimeters format
Format in my DB is:
4'6" (4 feet, 6 inches)
Formula for converting into centimeters
4*30.48 = 121.92 (convert feet to centimeters = multiply by 30.48)
6*2.54 = 15.24 (convert inches to centimeters = multiply by 2.54)
So Result = 121.92 + 15.24 = 137.16 cm
eg:
Actual Table: inches
SELECT * FROM inches
id height
1 4'6"
2 4'7"
3 5'8"
4 5'9"
I expect the following result as centimeters when I do SQL query
id height
1 137.16
2 139.7
3 172.72
4 175.26
Thanks in advance :)
Probably far easier to do on the application level, but if you really had to, you could do it in SQL like this, using the SUBSTR and INSTR functions, and some basic math:
SET #height = '4''6"';
SELECT
SUBSTR(#height, 1, INSTR(#height, '''') - 1) * 12 * 2.54 +
SUBSTR(#height, INSTR(#height, '''') + 1, INSTR(#height, '"') - INSTR(#height, '''') - 1) * 2.54;
-- yields 137.16
Or, applied to your table structure:
SELECT id,
SUBSTR(height, 1, INSTR(height, '''') - 1) * 12 * 2.54 +
SUBSTR(height, INSTR(height, '''') + 1, INSTR(height, '"') - INSTR(height, '''') - 1) * 2.54 AS height
FROM inches;
Application side process will be better,
However,
SELECT
(CAST(SUBSTR(height,1, LOCATE("'",height)-1) AS UNSIGNED) * 30.48) +
(CAST(SUBSTR(height, LOCATE("'",height)+1) AS UNSIGNED) * 2.54 ) AS cm
FROM
inches;
Can I ask for help on a SQL Statement please, I have to do the calculation inline and cannot declare variables for it
Calculation:
-91000000 / 2700000 = -33.7037037037
I need the remainder (7037037037 - but only up to 6 characters ) to be multiplied by 15000
703703 / 15000 = Final Answer of 49.913533
I thought I could do this:
select cast(ParseName(abs(cast(-91000000 as decimal)/ 2700000 ) %1,1) as numeric(8,8)) / 15000
WITH cte AS
(
SELECT -91000000 AS x, 2700000 AS y
)
SELECT ABS(ROUND((CAST(x AS decimal) / CAST(y AS decimal)) - (x/y), 6)) * 1000000 / 15000 FROM CTE