im using a v simple database and i have 3 columns A(bigINT 20) , B(bigInt 20) and c(DECIMAL(5,4)) , when i fire the following query i get the below mentioned results :
REPLACE INTO `my_table` SET `A` = 8,`B` = 44,`C` = 14;
i get these values in mysql A =8 , b= 44 and c as 9.9999 ! ?
any ideas as to why is this happening and what can i do to resolved this ?
DECIMAL(5,4) means that the number has at most 5 digits, 4 of them after decimal point. So 14 is simply overflow as it would require DECIMAL(6,4).
It must be cleared that 14 is overflow, because as constant precision point decimal it is internally 14.0000 here (so six digits over five).
So if you try to put 14.0000 (six digits) in DECIMAL(5,4) (five digits max) -> MySQL chooses value closest to the one you request. Therefore 14.0000 gets "rounded" to 9.9999.
To fit 14 in your column you can either extend it do DECIMAL(6,4) (to allow more digits in general) or change to DECIMAL(5,3) (which will allow one more digit before decimal point, but loses some precision of course).
Related
I get this error:
ERROR 1264 (22003): Out of range value for column 'median_comments' at
row 1
after running this query:
update influencers set `median_comments` = 1347 WHERE `id` = 1;
I'm not sure why this fails on this number which doesn't have any decimals and which is only 4 digits.
The field type is:
median_comments decimal(10,8)
You are using DECIMAL(10,8) that means max digits before decimal would be (10 - 8) = 2.
Reference: DECIMAL Data Type Characteristics
The declaration syntax for a DECIMAL column is DECIMAL(M,D). The ranges of values for the arguments are as follows:
M is the maximum number of digits (the precision). It has a range of 1 to 65.
D is the number of digits to the right of the decimal point (the scale). It has a range of 0 to 30 and must be no larger than M.
To fix the error, change your datatype to DECIMAL(10,2).
ALTER TABLE `influencers`
CHANGE COLUMN `median_comments` `median_comments` DECIMAL(10,2) NOT NULL DEFAULT 0;
If you are using decimal(10,8) as data type it means you are specifying 8 digits after decimal place which leaves only (10 - 8 i.e 2 digits) for your whole number.
In this case since your number 1347 contains 4 digits (whole number) hence you are getting the error as "Out of range value" since you are allowed only 2.
You should consider changing it to at least decimal (12,8) which will leave you 4 digits for your whole number part and your above command should work.
Please refer to post - Number format issue in Oracle. Same issue.
As you like me came here from google and your issue is related to Doctrine, and your column type is type="decimal", then you should configure precision and scale properties of your column in the right way.
For me, it was like before:
/** #ORM\Column(name="revenue", type="decimal", scale=4, nullable=true) */
private $revenue;
after
/** #ORM\Column(name="revenue", type="decimal", precision=14, scale=4, nullable=true) */
private $revenue;
It will be converted to DECIMAL(14,4), which means fourteen digits total, four of which are to the right of the decimal point.
Don't forget to prepare migration and run it to apply the changes.
Finally, you should get SQL like this:
ALTER TABLE project_finance CHANGE revenue revenue NUMERIC(14, 4) DEFAULT NULL
I want to know what's the best datatype to store these:
null
0
/* the length of other numbers is always 7 digits */
0000000
0000001
0000010
0000011
/* and so on */
1111111
I have tested, INT works as well. But there is a better datatype. Because all my numbers are made of 0 or 1 digits. Is there any better datatype?
What you are showing are binary numbers
0000000 = 0
0000001 = 2^0 = 1
0000010 = 2^1 = 2
0000011 = 2^0 + 2^1 = 3
So simply store these numbers in an integer data type (which is internally stored with bits as shown of course). You could use BIGINT for this, as recommended in the docs for bitwise operations (http://dev.mysql.com/doc/refman/5.7/en/bit-functions.html).
Here is how to set flag n:
UPDATE mytable
SET bitmask = POW(2, n-1)
WHERE id = 12345;
Here is how to add a flag:
UPDATE mytable
SET bitmask = bitmask | POW(2, n-1)
WHERE id = 12345;
Here is how to check a flag:
SELECT *
FROM mytable
WHERE bitmask & POW(2, n-1)
But as mentioned in the comments: In a relational database you usually use columns and tables to show attributes and relations rather than an encoded flag list.
As you've said in a comment, the values 01 and 1 should not be treated as equivalent (which rules out binary where they would be), so you could just store as a string.
It actually might be more efficient than storing as a byte + offset since that would take up 9 characters, whereas you need a maximum of 7 characters
Simply store as a varchar(7) or whatever the equivalent is in MySql. No need to be clever about it, especially since you are interested in extracting positional values.
Don't forget to bear in mind that this takes up a lot more storage than storing as a bit(7), since you are essentially storing 7 bytes (or whatever the storage unit is for each level of precision in a varchar), not 7 bits.
If that's not an issue then no need to over-engineer it.
You could convert the binary number to a string, with an additional byte to specify the number of leading zeros.
Example - the representation of 010:
The numeric value in hex is 0x02.
There is one leading zero, so the first byte is 0x01.
The result string is 0x01,0x02.
With the same method, 1010010 should be represented as 0x00,0x52.
Seems to me pretty efficient.
Not sure if it is the best datatype, but you may want to try BIT:
MySQL, PostgreSQL
There are also some useful bit functions in MySQL.
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
I have below table structure and data :
create table sample
(
id INT(10)
);
INSERT INTO sample
values
(23398),
(98743),
(54734);
Now I want to understand CAST function in mysql. Consider following query :
select
cast((id/3) as decimal(2,2)) as cast1,
cast((id/3) as decimal(3,2)) as cast2,
cast((id/3) as decimal(4,2)) as cast3,
cast((id/3) as decimal(5,2)) as cast4,
cast((id/3) as decimal(6,2)) as cast5,
cast((id/3) as decimal(7,2)) as cast6,
id/3 as actualId
from sample;
Please see output of this query at SQL Fiddle.
I am wondering why this query gives 0.99, 9.99 and vice versa.
Can anyone explain it ?
Thanks in advance.
decimal is a type that takes 2 arguments
decimal(size, places) :
size determines how many digits are in the number.
places determines how many of those digits are to the right of the decimal.
decimal(2,2) - .00 - 2 digits both of which are to the right of the decimal
when casting (23398 / 3) = 7799.33333333 to declimal(2, 2) it yields a decimal in the specified amount of space closest to the desired number which is 0.99
decimal(3,2) - 0.00 - 3 digits 2 of which are to the right of the decimal
when casting (23398 / 3) = 7799.33333333 to declimal(3, 2) it yields a decimal in the specified amount of space closest to the desired number which is 9.99
if all of the original numbers were negative you would yield -0.99 and -9.99 because they are the closest numbers to the desired number within the allocated space
As a matter of fact java does something similar if you take the max double and try to convert it to an int you will give the max int which is no where near the max double
SELECT COUNT(*) FROM area
WHERE ROUND(SQRT(POWER(('71' - coords_x), 2) +
POWER(('97' - coords_y), 2))) <= 17
==> 51
SELECT COUNT(*) FROM area
WHERE ROUND(SQRT(POWER((71 - coords_x), 2) +
POWER((97 - coords_y), 2))) <= 17
==> 22
coords_x and coords_y are both TINYINT fields containing values in the range [1, 150]. Usually MySQL doesn't care if numbers are quoted or not.. but apparently it does in this case.
The question is just: Why?
MySQL always cares about data types. What happens is that your code relies in automatic type casting and performs math on strings (which can hold a number or not). This can lead to all sort of unpredictable results:
SELECT POW('Hello', 'World') -- This returns 1
To sum up: you need to learn and use the different data types MySQL offers. Otherwise, your application will never do reliable calculations.
Update:
One more hint:
TINYINT[(M)] [UNSIGNED] [ZEROFILL]
A very small integer. The signed range
is -128 to 127. The unsigned range is
0 to 255.
URL:
http://dev.mysql.com/doc/refman/5.1/en/numeric-type-overview.html
I hope you are not trying to store 150 in a signed tinyint column.