Need to write a sql function that takes two different integers with the same length (same amount of digits) as arguments: one lower threshold and one upper threshold.
And the function should return a vector with all whole numbers between the two thresholds - but to reduce the array length, your function is supposed to return a wildcard character instead of the digits where possible.
Example: output of the function for a lower threshold L = 3778 and an upper threshold U = 9423
To further clarify, the line in the example showing 941* has one digit replaced by the wild card character and hence represents all values from 9410 - 9419
The line in the example showing 93* has two digits replaced by the wildcard character and represents all values from 9300 - 9399
And so on.
9423
9422
9421
9420
941*
93*
.
378*
3779
3778
I was checking out the difference between char vs varchar2 from google. I came across a word LEADING LENGTH in this link . THERE it was written that
Suppose you store the string ‘ORATABLE’ in a CHAR(20) field and a VARCHAR2(20) field. The CHAR field will use 22 bytes (2 bytes for leading length). The VARCHAR2 field will use 10 bytes only (8 for the string, 2 bytes for leading length).
Q1:How does the char field will use 22 bytes if the string is of 8 characters if (1 byte = 1 char)?
Q2 What is the LEADING LENGTH ? why it does occupy 2 bytes?
The CHAR() datatype pads the string with characters. So, for 'ORATABLE', it looks like:
'ORATABLE '
12345678901234567890
The "leading length" are two bytes at the beginning that specify the length of the string. Two bytes are needed because one byte is not enough. Two bytes allow lengths up to 65,535 units; one byte would only allow lengths up to 255.
The important point both CHAR() and VARCHAR2() use the same internal format, so there is little reason to sue CHAR(). Personally, I would only use it for fixed-length codes, such as ISO country codes or US social security numbers.
I'm trying to build an parser to deserialze into object. Socket will send byte into parser. For the length of field 22 POS Entry Mode will N3 and byte will be always 2 digit. How to get the value for this field ?
You read the ASCII value of this field, and convert it into an integer.
if it says N3 that means they are three digits numeric field, so if the value say 51, you cast it to 051 and send the ASCII equivalent
Field 22 is pos entry mode. It's 3 digit numeric value. If format is BCD then 2 bytes contains 4 digits[ 0 (padded) + 3 digit POS entry mode). If format is ascci then it is 3 byte.
I'm trying to write a query in oracle sql to get rows which has invalid 10 digit numbers, ie with other symbols in between them.
For example:
(111) 111-1111 #10 digit number with some symbols and spaces in between
111-111-1111
(111)111-1111
111)111-1111
(111) 11 1-1111
ie, It should match exactly 10 digit numbers which are non consecutive because it has some symbols in it.
So it should not match the following example:
111 #consecutive 3 digit number
11 1 #3 digit number with spaces
11-1 #3 digit number with symbol in between
1111111111 #consective 10 digit number
And I'm using REGEXP_LIKE, something like this
select * from table where REGEXP_LIKE(column, ?)
Any help is much appreciated. Thanks.
You could use a combination of a regex and length; the latter to exclude a pure 10-digit number without other characters:
regexp_like(col, '^[ .()-]*(\d[ .()-]*){10}$') and length(col) > 10
In the [.()-] class you would list all the characters that you would allow as symbols among the digits. Note that - needs to be the last in that list or else be escaped.
If you would allow any non-digit to occur among the 10 digits, you can use \D:
regexp_like(col, '^\D*(\d\D*){10}$') and length(col) > 10
So: the string should have length greater than 10, and the total number of digits must be exactly 10. This can be done without regular expressions (which should make it faster):
... where length(str) > 10 and
length(str) = 10 + length(translate(str, 'z0123456789', 'z'))
translate will translate the letter z to itself and all the other characters (digits) to nothing. Having to include the z is annoying, but unavoidable; translate will return NULL if any of its arguments is NULL. The second condition says the length of the input str is exactly 10 more than the length of the string with all digits removed - so there are exactly 10 digits.
I'm trying to emulate a function in SQL that a client has produced in Excel. In effect, they have a unique, 10-digit numeric value (VARCHAR) as the primary key in one of their enterprise database systems. Within another database, they require a unique, 5-digit alphanumeric identifier. They want that 5-digit alphanumeric value to be a representation of the 10-digit number. So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
The EXCEL equation is:
=IF(VALUE(MID(A2,1,4))>0,DEC2HEX(VALUE(MID(A2,3,2)))&DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX(VALUE(MID(A2,9,2))),DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX((VALUE(MID(A2,9,2)))))
I need the SQL equivalent of this. Of course, should someone out there know a better way to accomplish their goal of "a 5-digit alphanumeric identifier" based off the 10-digit number, I'm all ears.
ADDED 8/2/2011
First of all, thank you to everyone for the replies. Nice to see folks willing to help and even enjoying it! Based on all the responses, I'm apt to tell my client they're intent is sound, only their method is off kilter. I'd also like to recommend a solution. So the challenge remains, just modified slightly:
CHALLENGE: Within SQL, take a 10 digit, unique NUMERIC string and represent it ALPHANUMERICALLY in as few characters as possible. The resulting string must also be unique.
Note that the first 3-4 characters in the 10-digit string are likely to be zeros, and that they could be stripped to shorten the resulting alphanumeric string. Not required, but perhaps helpful.
This problem is inherently impossible. You have a 10 digit numeric value that you want to convert to a 5 digit alphanumeric value. Since there are 10 numeric characters, this means that there are 10^10 = 10 000 000 000 unique values for your 10 digit number. Since there are 36 alphanumeric characters (26 letters + 10 numbers), there are 36^5 = 60 466 176 unique values for your 5 digit number. You cannot map a set of 10 billion elements into a set with around 60 million.
Now, lets take a closer look at what your client's code is doing:
So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
This isn't 100% accurate. The excel code never uses the first 2 digits, but performs this operation on the remaining 8. There are two main problems with this algorithm which may not be intuitively obvious:
Two 10 digit numbers can map to the same 5 digit number. Consider the numbers 1000000117 and 1000001701. The last four digits of 1000000117 get mapped to 1 11, where the last four digits of 1000001701 get mapped to 11 1. This causes both to map to 00111.
The 5 digit number may not even end up being 5 digits! For example, 1000001616 gets mapped to 001010.
So, what is a possible solution? Well, if you don't care if that 5 digit number is unique or not, in MySQL you can use something like:
hex(<NUMERIC VALUE> % 0xFFFFF)
The log of 10^10 base 2 is 33.219280948874
> return math.log(10 ^ 10) / math.log(2)
33.219280948874
> = 2 ^ 33.21928
9999993422.9114
So, it takes 34 bits to represent this number. In hex this will take 34/4 = 8.5 characters, much more than 5.
> return math.log(10 ^ 10) / math.log(16)
8.3048202372184
The Excel macro is ignoring the first 4 (or 6) characters of the 10 character string.
You could try encoding in base 36 instead of 16. This will get you to 7 characters or less.
> return math.log(10 ^ 10) / math.log(36)
6.4254860446923
The popular base 64 encoding will get you to 6 characters
> return math.log(10 ^ 10) / math.log(64)
5.5365468248123
Even Ascii85 encoding won't get you down to 5.
> return math.log(10 ^ 10) / math.log(85)
5.1829075929158
You need base 100 to get to 5 characters
> return math.log(10 ^ 10) / math.log(100)
5
There aren't 100 printable ASCII characters, so this is not going to work, as zkhr explained as well, unless you're willing to go beyond ASCII.
I found your question interesting (although I don't claim to know the answer) - I googled a bit for you out of interest and found this which may help you http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html