Having this table, I would like to find the rows with Val fitting my Indata.
Tol field is a tolerance (varchar), that can be either an integer/float or a percentage value.
Row Val Tol Outdata
1 24 0 A
2 24 5 B
3 24 10 C
4 32 %10 D
5 32 1 E
Indata 30 for example should match rows 3 (24+10=34) and 4 (32-10%=28.8).
Can this be done in mySQL? CREATE FUNCTION?
This is going to be rather difficult to do in MySQL with that table and column design. How do you plan to differentiate what sort of comparison should be done? By doing a string comparison to see if your varchar field contains a percentage sign?
I would suggest breaking your tolerance field into (at least) two int/float columns, say tol and tol_pct. For flexibility, I would represent tol_pct as a decimal (10% => .10). Then, you can do a query that looks like:
select *
from table
where
(Indata between Val - tol and Val + tol)
or (Indata between Val * (1 + tol_pct) and Val * (1 - tol_pct))
I don't have a MySQL install to test it on, but this example is converted from Oracle sql syntax. You have to use string functions to determine if the tol is a percent and act accordingly to calculate the min and max range for that field. Then you can use a between clause.
select *
from (select t.*,
case when substr(tol, 1, 1) = '%' then
t.val * (1 + convert('.' + substr(tol, 2), number))
else
t.val + convert(tol, number)
end maxval,
case when substr(tol, 1, 1) = '%' then
t.val * (1 - convert('.' + substr(tol, 2), number))
else convert(t.val - tol, number)
end minval
from mytable
) t
where 30 between minval and maxval
;
Sure it can be done. But let me tell you that you better review your database design as it conflicts with normalization quite a bit. See http://en.wikipedia.org/wiki/Database_normalization for a quick overview.
Related
In SQL there are aggregation operators, like AVG, SUM, COUNT. Why doesn't it have an operator for multiplication? "MUL" or something.
I was wondering, does it exist for Oracle, MSSQL, MySQL ? If not is there a workaround that would give this behaviour?
By MUL do you mean progressive multiplication of values?
Even with 100 rows of some small size (say 10s), your MUL(column) is going to overflow any data type! With such a high probability of mis/ab-use, and very limited scope for use, it does not need to be a SQL Standard. As others have shown there are mathematical ways of working it out, just as there are many many ways to do tricky calculations in SQL just using standard (and common-use) methods.
Sample data:
Column
1
2
4
8
COUNT : 4 items (1 for each non-null)
SUM : 1 + 2 + 4 + 8 = 15
AVG : 3.75 (SUM/COUNT)
MUL : 1 x 2 x 4 x 8 ? ( =64 )
For completeness, the Oracle, MSSQL, MySQL core implementations *
Oracle : EXP(SUM(LN(column))) or POWER(N,SUM(LOG(column, N)))
MSSQL : EXP(SUM(LOG(column))) or POWER(N,SUM(LOG(column)/LOG(N)))
MySQL : EXP(SUM(LOG(column))) or POW(N,SUM(LOG(N,column)))
Care when using EXP/LOG in SQL Server, watch the return type http://msdn.microsoft.com/en-us/library/ms187592.aspx
The POWER form allows for larger numbers (using bases larger than Euler's number), and in cases where the result grows too large to turn it back using POWER, you can return just the logarithmic value and calculate the actual number outside of the SQL query
* LOG(0) and LOG(-ve) are undefined. The below shows only how to handle this in SQL Server. Equivalents can be found for the other SQL flavours, using the same concept
create table MUL(data int)
insert MUL select 1 yourColumn union all
select 2 union all
select 4 union all
select 8 union all
select -2 union all
select 0
select CASE WHEN MIN(abs(data)) = 0 then 0 ELSE
EXP(SUM(Log(abs(nullif(data,0))))) -- the base mathematics
* round(0.5-count(nullif(sign(sign(data)+0.5),1))%2,0) -- pairs up negatives
END
from MUL
Ingredients:
taking the abs() of data, if the min is 0, multiplying by whatever else is futile, the result is 0
When data is 0, NULLIF converts it to null. The abs(), log() both return null, causing it to be precluded from sum()
If data is not 0, abs allows us to multiple a negative number using the LOG method - we will keep track of the negativity elsewhere
Working out the final sign
sign(data) returns 1 for >0, 0 for 0 and -1 for <0.
We add another 0.5 and take the sign() again, so we have now classified 0 and 1 both as 1, and only -1 as -1.
again use NULLIF to remove from COUNT() the 1's, since we only need to count up the negatives.
% 2 against the count() of negative numbers returns either
--> 1 if there is an odd number of negative numbers
--> 0 if there is an even number of negative numbers
more mathematical tricks: we take 1 or 0 off 0.5, so that the above becomes
--> (0.5-1=-0.5=>round to -1) if there is an odd number of negative numbers
--> (0.5-0= 0.5=>round to 1) if there is an even number of negative numbers
we multiple this final 1/-1 against the SUM-PRODUCT value for the real result
No, but you can use Mathematics :)
if yourColumn is always bigger than zero:
select EXP(SUM(LOG(yourColumn))) As ColumnProduct from yourTable
I see an Oracle answer is still missing, so here it is:
SQL> with yourTable as
2 ( select 1 yourColumn from dual union all
3 select 2 from dual union all
4 select 4 from dual union all
5 select 8 from dual
6 )
7 select EXP(SUM(LN(yourColumn))) As ColumnProduct from yourTable
8 /
COLUMNPRODUCT
-------------
64
1 row selected.
Regards,
Rob.
With PostgreSQL, you can create your own aggregate functions, see http://www.postgresql.org/docs/8.2/interactive/sql-createaggregate.html
To create an aggregate function on MySQL, you'll need to build an .so (linux) or .dll (windows) file. An example is shown here: http://www.codeproject.com/KB/database/mygroupconcat.aspx
I'm not sure about mssql and oracle, but i bet they have options to create custom aggregates as well.
You'll break any datatype fairly quickly as numbers mount up.
Using LOG/EXP is tricky because of numbers <= 0 that will fail when using LOG. I wrote a solution in this question that deals with this
Using CTE in MS SQL:
CREATE TABLE Foo(Id int, Val int)
INSERT INTO Foo VALUES(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)
;WITH cte AS
(
SELECT Id, Val AS Multiply, row_number() over (order by Id) as rn
FROM Foo
WHERE Id=1
UNION ALL
SELECT ff.Id, cte.multiply*ff.Val as multiply, ff.rn FROM
(SELECT f.Id, f.Val, (row_number() over (order by f.Id)) as rn
FROM Foo f) ff
INNER JOIN cte
ON ff.rn -1= cte.rn
)
SELECT * FROM cte
Not sure about Oracle or sql-server, but in MySQL you can just use * like you normally would.
mysql> select count(id), count(id)*10 from tablename;
+-----------+--------------+
| count(id) | count(id)*10 |
+-----------+--------------+
| 961 | 9610 |
+-----------+--------------+
1 row in set (0.00 sec)
In Oracle 11g, I am trying to get to a sell price from a query of data. Yes I can export this and write the code somewhere else, but I want to try to do this elegantly in the query.
I only seem to get the first part of the equation and not the last CASE where I use:
WHEN sales_code
What I am ultimately trying to do is take the result from the top and divide it by the bottom except in the case of SALE_CODE 4 where I add 1+1 or 2 to the top result and then divide by the equation.
round(to_number(price) *
CASE WHEN class_code='X'
THEN .48
ELSE .5
END * e1.set_qty +
CASE WHEN carton_pack_qty = '1'
THEN 0
ELSE (
CASE WHEN NVL(SUBSTR(size, 1,NVL(LENGTH(size) - 2,0)),1) > '35'
THEN 3.5
ELSE 3
END)
END +
CASE
WHEN sales_code='1' THEN 0 /(1-17/100)
WHEN sales_code='2' THEN 0 /(1-5/100)
WHEN sales_code='3' THEN 0 /(1-18/100)
WHEN sales_code='4' THEN 1+1 / (1-9.5/100)
WHEN sales_code='5' THEN 0 /(1-17/100)
WHEN sales_code='6' THEN 0 /(1-8/100)
WHEN sales_code='7' THEN 0 /((1-150)/100)
ELSE (100/100)
END,2) AS "Price",
I get a result from the query, but not the whole calculation. I tried this many other ways and there was always an error with parentheses or some other arbitrary error.
Any help would be appreciated.
I think this is your problem:
WHEN sales_code='1' THEN 0 /(1-17/100)
CASE returns a scalar, a number. You're trying to have it return the second half of the formula in your calculation. You need something more like this:
...
END +
CASE WHEN sales_code='4' THEN 1 ELSE 0 END /
CASE
WHEN sales_code='1' THEN (1-17/100)
WHEN sales_code='2' THEN (1-5/100)
WHEN sales_code='3' THEN (1-18/100)
WHEN sales_code='4' THEN (1-9.5/100)
WHEN sales_code='5' THEN (1-17/100)
WHEN sales_code='6' THEN (1-8/100)
WHEN sales_code='7' THEN ((1-150)/100)
ELSE 1 END ...
Actually, I'm not entirely sure what you're trying to do with sales_code='4', but that looks close.
I think I understand now what you are trying to do. Almost at least :-)
The first thing you should do is write down the complete formula with parentheses where needed. Something like:
final = ((price * class_code_factor * set_qty) + quantity_summand + two_if_sales_code4) * sales_code_factor
(That last part looks like a percentage factor, not a divisor to me. I may be wrong of course.)
Once you have the formula right, translate this to SQL:
ROUND
(
(
(
TO_NUMBER(price) *
CASE WHEN class_code = 'X' THEN 0.48 ELSE 0.5 END *
e1.set_qty
)
+
CASE WHEN carton_pack_qty = 1 THEN 0
ELSE CASE WHEN NVL(SUBSTR(size, 1,NVL(LENGTH(size) - 2,0)),1) > '35'
THEN 3.5
ELSE 3
END
END
+
CASE WHEN sales_code = 4 THEN 2 ELSE 0 END
)
*
CASE
WHEN sales_code = 1 THEN 1 - (17 / 100)
WHEN sales_code = 2 THEN 1 - (5 / 100)
WHEN sales_code = 3 THEN 1 - (18 / 100)
WHEN sales_code = 4 THEN 1 - (9.5 / 100)
WHEN sales_code = 5 THEN 1 - (17 / 100)
WHEN sales_code = 6 THEN 1 - (8 / 100)
WHEN sales_code = 7 THEN (1 - 150) / 100)
ELSE 1
END
, 2 ) AS "Price",
Adjust this to the formula you actually want. There are some things I want to point out:
Why is price not a number in your database, but a string that you must convert to a number with TO_NUMBER? That must not be. Store values in the appropriate format in your database.
In a good database you would not have to get a substring of size. It seems you are storing two different things in this column, which violates database normalization. Separate the two things and store them in separate columns.
The substring thing looks strange at that, too. You are taking the left part of the size leaving out the last two characters. It seems hence that you don't know the lenth of the part you are getting, so let's say that this can be one, two or three characers. (I don't know of course.) Now you compare this result with another string; a string that contains a numeric value. But as you are comparing strings, '4' is greater than '35', because '4' > '3'. And '200' is lesser than '35' because '2' < '3'. Is this really intended?
There are more things you treated as strings and I took the liberty to change this to numbers. It seems for instance that a quantity (carton_pack_qty) should be stored as a number. So do this and don't compare it to the string '1', but to the number 1. The sales code seems to be numeric, too. Well, again, I may be wrong.
In a good database there would be no magic numbers in the query. Knowledge belongs in the database, not in the query. If a class code 'X' means a factor of 0.48 and other class codes mean a factor of 0.5, then why is there no table of class codes showing what a class code represents and what factor to apply? Same for the mysterious summand 3 resp. 3.5; there should be a table holding these values and the size and quantity ranges they apply to. And at last there is the sales code which should also be stored in a table showing the summand (2 for code 4, 0 elsewise) and the factor.
The query part would then look something like this:
ROUND((price * cc.factor * el.set_qty) + qs.value + sc.value) * sc.factor, 2) AS "Price"
Breaking the dividend into a sub query worked and then adding parentheses around it to divide by in the main query worked.
(
select
style,
to_number(price) *
CASE WHEN class_code='X'
THEN .48
ELSE .5
END * set_qty +
CASE WHEN carton_pack_qty = '1'
THEN 1
ELSE (
CASE WHEN to_number(NVL(SUBSTR(size, 1,NVL(LENGTH(size) - 2,0)),1)) > 35
THEN 3.5
ELSE 3
END)
END as Price
FROM STYL1 s1,STY2 s2
WHERE s1.style=s2.style
) P1
From the data in the first table how do I get a simple percentage of total calculation:
two / (two + three) = 2 / (2 + 3) = 0.4
to arrive at:
You can try the below -
select ((case when name='two' then value end)*1.0)/sum(value)
from t
where name in ('two','three')
You would rarely do something like this in SQL. Anyway, you can do:
with
two as (select value from t where name = 'two'),
three as (select value from t where name = 'three')
select 1.0 * two.value / (two.value + three.value) as answer
from two
cross join three
Result:
answer
----------------------
0.40000000000000000000
See running example at DB Fiddle.
I came across the following SQL statement:
SELECT A.NAME
FROM THE_TABLE A
WHERE A.NAME LIKE '%JOHN%DOE%'
AND ((A.NUM_FIELD/1) - (A.NUM_FIELD/2)*2 <> 0)
That last condition, "((A.NUM_FIELD/1) - (A.NUM_FIELD/2)*2 <> 0)" is what baffles me. Depening on the implementation of order of operations, it should always result to 0 or A.NUM_FIELD / 2.
How does SQL still return records from this view? If it always results to half the original value, why have it? (This is a delivered SQL package)
Probably integer division, so an odd NUM_FIELD is going to be one less.
MSDN says:
If an integer dividend is divided by an integer divisor, the result is
an integer that has any fractional part of the result truncated.
if the NUM_FIELD is an integer, and an odd one- then
(A.NUM_FIELD/1) - (A.NUM_FIELD/2)*2
is equal to one
What SQL implementation is this?
As noted,
(x/1) - (x/2)*2
is equivalent to
X - (2*(x/2))
which, if integer division is being performed yields 0 or 1 depending on whether the value is even or odd:
x x/2 2*(x/2) x-(2*(x/2))
- --- ------- -----------
0 0 0 0
1 0 0 1
2 1 2 0
3 1 2 1
4 2 4 0
...
if so, it seems like an odd way way of checking for odd/even values, especially since most SQL implementations that I'm aware of support a modulo operator or function, either x % y or mod(x,y).
The seemingly extraneous division by 1 makes me think the column in question might be floating point. Perhaps the person who coded the query is trying to check for jitter or fuzzyness in the low order bits of the floating point column?
if you modified the query to return all the intermediate values of the computation:
SELECT A.NAME as Name ,
A.NUM_FIELD as X ,
A.NUM_FIELD / 1 as X_over_1 ,
A.NUM_FIELD / 2 as X_over_2 ,
( X.NUM_FIELD / 2 )
* 2 as 2x_over_2 ,
( A.NUM_FIELD / 1 )
- ( A.NUM_FIELD / 2 ) * 2 as Delta ,
case when ( ( A.NUM_FIELD / 1 ) - ( A.NUM_FIELD / 2 ) * 2 ) <> 0
then 'return'
else 'discard'
end as Row_Status
FROM THE_TABLE A
WHERE A.NAME LIKE '%JOHN%DOE%'
and then executed it, what results do you get?
I was working on a query today which required me to use the following to find all odd number ID values
(ID % 2) <> 0
Can anyone tell me what this is doing? It worked, which is great, but I'd like to know why.
ID % 2 is checking what the remainder is if you divide ID by 2. If you divide an even number by 2 it will always have a remainder of 0. Any other number (odd) will result in a non-zero value. Which is what is checking for.
For finding the even number we should use
select num from table where ( num % 2 ) = 0
As Below Doc specify
dividend % divisor
Returns the remainder of one number divided by another.
https://learn.microsoft.com/en-us/sql/t-sql/language-elements/modulo-transact-sql#syntax
For Example
13 % 2 return 1
Next part is <> which denotes Not equals.
Therefor what your statement mean is
Remainder of ID when it divided by 2 not equals to 0
Be careful because this is not going to work in Oracle database. Same Expression will be like below.
MOD(ID, 2) <> 0
ID % 2 reduces all integer (monetary and numeric are allowed, too) numbers to 0 and 1 effectively.
Read about the modulo operator in the manual.
In oracle,
select num from table where MOD (num, 2) = 0;
dividend % divisor
Dividend is the numeric expression to divide. Dividend must be any expression of integer data type in sql server.
Divisor is the numeric expression to divide the dividend. Divisor must be expression of integer data type except in sql server.
SELECT 15 % 2
Output
1
Dividend = 15
Divisor = 2
Let's say you wanted to query
Query a list of CITY names from STATION with even ID numbers only.
Schema structure for STATION:
ID Number
CITY varchar
STATE varchar
select CITY from STATION as st where st.id % 2 = 0
Will fetch the even set of records
In order to fetch the odd records with Id as odd number.
select CITY from STATION as st where st.id % 2 <> 0
% function reduces the value to either 0 or 1
It's taking the ID , dividing it by 2 and checking if the remainder is not zero; meaning, it's an odd ID.
<> means not equal. however, in some versions of SQL, you can write !=