I am working on an application that requires me to validate if 3 randomly generated numbers match a 3 digit string that has been entered into a database from user input. I also need to preserve the exact order that the user enters the string, so sorting on input is not an option.
For example, the randomly generated digits may be 6 4 0, and in the database a string may show as '406'.
Is there an easy way this can be accomplished in a single query without enumerating the options or adding an extra column/view?
maybe you could try
create table y (z varchar(10));
insert into y values ('406');
insert into y values ('604');
insert into y values ('446');
insert into y values ('106');
insert into y values ('123');
and then
SELECT * from y where FIND_IN_SET(Substring('640',1,1),MAKE_SET(7,Substring(z,1,1),Substring(z,2,1),Substring(z,3,1))) and FIND_IN_SET(Substring('640',2,1),MAKE_SET(7,Substring(z,1,1),Substring(z,2,1),Substring(z,3,1))) and FIND_IN_SET(Substring('640',3,1),MAKE_SET(7,Substring(z,1,1),Substring(z,2,1),Substring(z,3,1)));
returns
406
604
Sum the three random digits
Something like
Select * From Triplets Where (Ascii(Substring(Number,0,1)) - 48) + (Ascii(substring(Number,1,1)) -48) +
(Ascii(substring(Number,2,1)) -48) = MySumOfNumber
easy is a state of mind isn't it, Storage requirement of an extra "CheckSum" int, versus the high cost of a query like this.
Related
I have a data set like that:
Data Set Contents
From To Comment
----+---+--------
0 50 Bad
50 70 Good
70 100 Excellent
If I have a value of 75, I need to get Excellent by searching the Dataset.
I know about the lookup function but it is not what I want. How can I do that?
The values should be in percentage.
Note : the value (75) is Average of a column (Calculated) it
calculate student grade from max and student mark Version SQL Server
2016
Note 2 : the dataset is from database not static values
Thank You
Assuming you only ever have a fixed number of 'grades' then this will work. However, I would strongly recommend doing this type of work on the server where possible.
Here we go...
I created two datasets
dsGradeRange with the following sql to recreate your example (more or less)
DECLARE #t TABLE (low int, high int, comment varchar(20))
INSERT INTO #t VALUES
(0,49,'Bad'),
(50,69,'Good'),
(70,100, 'Excellent')
SELECT * FROM #t
dsRandomNumbers This just creates 30 random numbers between 0 and 100
SELECT *
FROM (SELECT top 30 ABS(CHECKSUM(NEWID()) % 100) as myNumber FROM sys.objects) x
ORDER BY myNumber
I added a table to the report to show the grades (just for reference).
I then added a table to show the dsRandomNumbers
Finally I set the expression of the 2nd column to the following expression.
=SWITCH
(
Fields!myNumber.Value < LOOKUP("Bad", Fields!comment.Value, Fields!high.Value, "dsGradeRange"), "Bad",
Fields!myNumber.Value < LOOKUP("Good", Fields!comment.Value, Fields!high.Value, "dsGradeRange"), "Good",
True, "Excellent"
)
This gives the following results
As you can see we only need to compare to the high value of each case, the first match will return the correct comment.
Right click on your dataset and add a calculated field. Go to Field Properties > Fields > Add and add the following expression, which descripes your scenario:
=IIF(Fields!Number.Value < 50, "Bad", "Good")
I need round up the numbers of a complete table of mysql with a funtion leaving the two decimals to zero?
e.g.:
from 75,55 to 86,00
from 75,45 to 75,00
All the prices of a complete table changed a few decimal and I do not know how it happened.
Can anybody help me?
You can simply use Round(X, D) function. From Docs:
Rounds the argument X to D decimal places. The rounding algorithm
depends on the data type of X. D defaults to 0 if not specified. D can
be negative to cause D digits left of the decimal point of the value X
to become zero.
Try:
SELECT ROUND(75.45, 2);
-- Fetches 75.00
SELECT ROUND(75.55, 2);
-- Fetches 76.00
Now, it seems that you want to Update all the values of a particular column, by rounding them to two decimal places. You can do the following:
UPDATE your_table_name
SET your_column_name = ROUND(your_column_name, 2);
-- It will round all the values to 2 decimal places.
"UPDATE your_table_name SET your_column_name = ROUND(your_column_name, 2);" not work for me. I dont know.
¡SOLUTION! = TABLE > ESTRUCTURE > NAME COLUMN > length/values > remove decimal.
All number round up in my row (column). Them I put decimal again. All my fields it's ok.
When I define a table to store decimal values I use a statement like this:
CREATE TABLE myTable (
myKey INT NOT NULL,
myValue DECIMAL(10,2) NOT NULL,
PRIMARY KEY (myKey)
);
However, this results in every myValue being stored with a one-size-fits-all precision of (10,2). For instance
45.6 becomes 45.60
21 becomes 21.00
17.008 becomes 17.01
But what if each record has a myValue of different precision? I need 45.6 to remain 45.6, 21 to remain 21, and 17.008 to remain 17.008. Otherwise the precision of measurement is being lost. There's a big difference between 21 and 21.00.
If you don't need to do greater/less-than compares, store as a VARCHAR(..)
The strings '21' and '21.00' would have identical values, but present different "precision".
When needing the numeric value, add zero (col + 0).
This does not allow for "negative precision", such as "1.2M" being represented as 1200000. If you need that, then Norbert's approach is probably better.
You can store with high precision and exact recall by following a different way of storing the data:
Create a table with two columns:
CREATE TABLE precise (value BIGINT, decimaldot INT);
Use code to determine where the dot is, for example in your 21 value: 2 (assuming 1 indexing). So stored the value would be:
INSERT INTO precise values (21,2);
Retrieved it would return 21 exact (parsing back the dot in the value 21 at position 2, is 21)
Value 17.008 would also have decimaldot at 2:
INSERT INTO precise values (17008,2);
Etc..
Larger values can be stored by using a VARCHAR(4000) instead of a biginteger, or by using blob fields.
Hey is there any way to create query with simple formula ?
I have a table data with two columns value_one and value_two both are decimal values. I want to select this rows where difference between value_one and value_two is grater then 5. How can i do this?
Can i do something like this ?
SELECT * FROM data WHERE (MAX(value_one, value_two) - MIN(value_one, value_two)) > 5
Example values
value_one, value_two
1,6
9,3
2,3
3,2
so analogical difs are: 5, 6, 1, 1 so the selected row would be only first and second.
Consider an example where smaller number is subtracted with a bigger number:
2 - 5 = -3
So, the result is a difference of two numbers with a negation sign.
Now, consider the reverse scenario, when bigger number is subtracted with the smaller number:
5 - 2 = 3
Pretty simple right.
Basically, the difference of two number remains same, if you just ignore the sign. This is in other words called absolute value of a number.
Now, the question arises how to find the absolute value in MySQL?
Answer to this is the built-in method of MySQL i.e. abs() function which returns an absolute value of a number.
ABS(X):
Returns the absolute value of X.
mysql> SELECT ABS(2);
-> 2
mysql> SELECT ABS(-32);
-> 32
Therefore, without worrying about finding min and max number, we can directly focus on the difference of two numbers and then, retrieving the absolute value of the result. Finally, check if it is greater than 5.
So, the final query becomes:
SELECT *
FROM data
WHERE abs(value_one - value_two) > 5;
You can also do complex operations once the absolute value is calculated like adding or dividing with the third value. Check the code below:
SELECT *
FROM
data
WHERE
(abs(value_one - value_two) / value_three) + value_four > 5;
You can also add multiple conditions using logical operators like AND, OR, NOT to do so. Click here for logical operators.
SELECT *
FROM
data
WHERE
((abs(value_one - value_two) / value_three) + value_four > 5)
AND (value_five != 0);
Here is the link with various functions available in MySQL:
https://dev.mysql.com/doc/refman/5.0/en/mathematical-functions.html
No, you would just use a simple where clause:
select *
from data
where abs(value_one - value_two) > 5;
I have a field in the mysql database that contains data like the following:
Q16
Q32
L16
Q4
L32
L64
Q64
Q8
L1
L4
Q1
And so forth. What I'm trying to do is pull out, let's say, all the values that start with Q which is easy:
field_name LIKE 'Q%'
But then I want to filter let's say all the values that have a number higher than 32. As a result I'm supposed to get only 'Q64', however, I also get Q4, Q8 and so for as I'm comparing them as strings so only 3 and the respective digit are compared and the numbers are in general taken as single digits, not as integers.
As this makes perfect sense, I'm struggling to find a solution on how to perform this operation without pulling all the data out of the database, stripping out the Qs and parsing it all to integers.
I did play around with the CAST operator, however, it only works if the value is stored as string AND it contains only digits. The parsing fails if there's another character in there..
Extract the number from the string and cast it to a number with *1 or cast
select * from your_table
where substring(field_name, 1, 1) = 'Q'
and substring(field_name, 2) * 1 > 32