Basically what I'm trying to do is add '000' to a number (between 5-8 characters in length) and make the whole numbers have decimals.
What I came up with is:
SELECT DISTINCT
'000' || TO_CHAR(Blah, '9,999,999.99') AS "Data"
FROM Blah database
While this does what I ideally want, there is a gap between the zeroes of either 3 or 4 depending on the number. Obviously I don't want the gap there. Where am I going astray?
Use trim(to_char(x, '9,999,999.99')) this way you will avoid gap
Related
How do we convert and add trailing and leading zeros to a number? For example 123.45. I need to make this ten digits long and have padding numbers in front and back. I would like to convert it to 0001234500. Two trailing numbers after the last digit of the decimal. Remove the decimal. Fill in the remaining space with zeroes for the leading end.
I have this so far and it adds trailing zeroes and removes the decimal.
REPLACE(RIGHT('0'+CAST(rtrim(convert(char(10),convert(decimal(10,4),Field))) AS VARCHAR(10)),10),'.','') as New_Field
In MySQL you would have RPAD and LPAD to get stuff like this done, in SQL Server (2012+) you can get something similar by working with FORMAT.
Easiest way is to FORMAT your numbers with a dot so that they take the right place in the format string, then remove that dot. You need to specify a locale, since in different regions you will get a different decimal sign (even if you use . within the format pattern, you would get , in various locales) - using en-US makes sure you get a dot.
REPLACE(FORMAT(somenumber, '000000.0000', 'en-US'), '.', '')
A few examples:
WITH TempTable(somenumber) AS (
SELECT 3
UNION SELECT 3.4
UNION SELECT 3.45
UNION SELECT 23.45
UNION SELECT 123.45
)
SELECT
somenumber,
REPLACE(FORMAT(somenumber, '000000.0000', 'en-US'), '.', '')
FROM
TempTable;
Gives
3.00 0000030000
3.40 0000034000
3.45 0000034500
23.45 0000234500
123.45 0001234500
You seem to really be overthinking what you need to do here. If we take it in steps, perhaps you'll see that this can be achieved much more easily. This solution runs under the idea that the value 123.45 becomes 0001234500 and 6.5 becomes 0000065000.
Firstly, let's pad out the right hand side of the number 123.45 so that we have 1234500 That's easy enough : 123. 45 * 100 = 12345 So, to get 1234500 we simply need to multiple it by a couple of extra factors of 10:
SELECT 123.45 * 10000; --1234500.00
Ok, now, let's get rid of those decimal places. Easiest way, convert it to an int:
SELECT CONVERT(int, 123.45 * 10000); --1234500
Nice! Now, the finalstep, the leading 0's. A numerical value, in SQL Server, won't display leading zeros. SELECT 01, 001.00; Will return 1 and 1.00 respectively. A varchar however, will though (as it's not a number). We can, therefore, make use of that with a further conversion, and then then use of RIGHT:
SELECT RIGHT('0000000000' + CONVERT(varchar(10),CONVERT(int,123.45 * 10000)),10);
Now you have the value you want '0001234500'.
If you're only after padding, (so 6.5 becomes 0006500) then you should be able to work out how to achieve this with the help above (hint you don't need RIGHT).
Any questions, please do ask.
I have a requirement to convert the very lengthy amount to a comma separated value in oracle. I was searching in google. but I got some solutions which works only for small numbers. But not for lengthy number. Below is the solution I have. But not working properly. I was getting ############... if I run the below.
SELECT TO_CHAR(6965854565787645667634565432234565432345643265432345643242087,
'99G999G999G9999', 'NLS_NUMERIC_CHARACTERS=",."') as test
FROM dual;
Desired output:
6,965,854,565,787,645,667,634,565,432,234,565,432,345,643,265,432,345,643,242,087
Please help me. thanks in advance.
Please check if below query can help.
SELECT ltrim(regexp_replace('00'
|| '6965854565787645667634565432234565432345643265432345643242087', '(...)', ',\1' ),',|0') AS t
FROM dual;
Numbers in Oracle can't have more than 38 significant digits. You have many more than that.
If I may, what kind of "amount" is that? My understanding is that Oracle was designed to handle real-life values. What possible meaning is there to the sample number you posted?
Added: Original poster in a comment (below) stated that he is getting the same error with a shorter number, only 34 digits.
Two issues. First, the format model must have at least the needed number of digits (of 9's). to_char(100000, '9G999') will produce the output #### because the format model allows only 4 digits, but the input is 6 digits.
Then, after that is corrected, the output may still look incorrect in the front-end application, like SQL*Plus. In SQL*Plus the default width of a number column is 10 (I believe). That can be changed to 38, for example with the command set numwidth 38. In other front-ends, like Toad and SQL Developer, the default numeric width is a setting that can be changed through the graphical user interface.
More added - actually the result of to_char is a string, and by default strings of any length should be displayed OK in any front-end, so the numeric width is probably irrelevant. (And, in any case, it does not affect the displaying of strings, including the result of to_char().)
SELECT TO_CHAR(
6676345654322345654323456432654323456,
'999G999G999G999G999G999G999G999G999G999G999G999G999',
'NLS_NUMERIC_CHARACTERS=",."') as test FROM dual
TEST
------------------------------------------------------------
6,676,345,654,322,345,654,323,456,432,654,323,456
#AlexPoole pointed out that perhaps your input is a string.
I didn't get that vibe; but if in fact your input IS a string, and if you know the length is no more than 99 digits, you could do something like below. If your strings can be longer than 99, replace 99 below with a sufficiently large multiple of 3. (Or, you can replace it with a calculated value, 3 * ceil(length(str)/3)).
with
inputs ( str ) as (
select '12345678912345' from dual
)
-- WITH clause is only for testing/illustration, not part of the solution
select ltrim(regexp_replace(lpad(str, 99, ','), '(.{3})', ',\1'), ',') as test
from inputs;
TEST
------------------
12,345,678,912,345
Can someone please finalize the code on the below.
I only want to look for a 6 digit number range anywhere in the RMK field, between 100000 and 999999
REGEXP_LIKE(RMKADH.RMK, '[[:digit:]]')
The current code works but is bringing back anything with a number so I'm trying to narrow it down to 6 digits together. I've tried a few but no luck.
Edit:
I want to flag this field if a 6 digit number is present. The reference will always be 6 digits long only, no more no less. But as it's a free text field it could be anywhere and contain anything.
Example output I do want to flag: >abc123456markj< = flagged.
Output I don't want to flag: >Mark 23647282< because the number it finds is more than 6 characters in length I know it's not a valid reference.
Try this:
REGEXP_LIKE(RMKADH.RMK, '[1-9][[:digit:]]{5}') AND length(RMKADH.RMK) = 6
For more info, see: Multilingual Regular Expression Syntax
You can do a REGEXP_SUBSTR to get 6 digits out of the given field and compare it using between
select * from t
where to_number(regexp_substr(col,'[[:digit:]]{6}')) between 100000 and 999999;
;
Please note that if a bigger sequence than 6 digits exists, the above solution will take first 6 digits into consideration. If you want to do for any 6 consecutive digits, the solution will have to be a different one.
If you want to get all the Records which have only Numeric values in them you can use below query
REGEXP_LIKE(RMKADH.RMK, '^[[:digit:]]+$');
The above will match any number of Numbers from start to end in the string. So if your Numbers span from 1 digit to any number of Digits, this will be useful.
SELECT
to_number(regexp_replace('abc123456markj', '[^[:digit:]]', '')) digits
FROM
dual
WHERE
REGEXP_LIKE('abc123456markj', '[[:digit:]]')
AND
length(regexp_replace('abc123456markj', '[^[:digit:]]', '')) = 6
AND
regexp_replace('abc123456markj', '[^[:digit:]]', '') BETWEEN 100000 AND 999999;
I have a column of numbers between 0 - 6 digits long. For those less than 6 I need to pad out with zeros to ensure they are all 6 digits i.e 12563 = 012563 or 23 000023 etc etc. Can someone recommend a solution?
Probably the easiest way to pad numbers with leading zeros would be to use the Format() function, as in
Format(fieldName, "000000")
If you're searching on this (like for PIN numbers, where '12' would be represented as '000012' here's an example using Gord's correct answer;
SELECT CStr(Format(fieldName,"000000")) FROM table WHERE CStr(Format(fieldName,"000000"))="000012";
I had a similar issue. I couldn't change the field on the actual file because it was a split database and it had to be changed on the data source (Database_be). I went to the data source and made the change from Number to Short Text to all tables and that was it... Like magic!!
Try:
Update TABLE set DIGITS = string(6- len(DIGITS),"0")
DIGITS TABLE is the table where your numbers are stored.
DIGITS is the field that contains your numbers.
The above does NOT work.
Corrected version:
Update TABLE set DIGITS = string(6- len(DIGITS),"0")&DIGITS
The number '6' can be altered for whatever the total length of your field.
I'm currently battling with something
that must be trivial for you.
I have 2 number 191 and 97, and I need to put them in a SQL request, as chars and 97 must be printed as 097.
At first I tried 999, but it added 2 space to my numbers.
then 099, it does print 097 but it adds a space to it.
to_char(:center, '099') = " 197" and " 097"
Where is this space coming from?
Thanks.
What you're looking for is the Format Modifier element:
to_char(:center, 'fm099')
The leading space is for the potential minus sign. To remove it you can use FM in the format:
to_char(v_num,'FM099')
9 9999 Returns value with the specified number of digits with a leading space if positive or with a leading minus if negative.Leading zeros are blank, except for a zero value, which returns a zero for the integer part of the fixed-point number.
From http://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements004.htm#i34510
Use #DavidAldridge solution, or trim your value.
If you are looking for the all column values in same number of digits even the actual value having less digits. Try this
a) SELECT TO_CHAR(COLUMN_NAME, 'FM099') FROM TABLE_NAME;
b) SELECT TO_CHAR(COLUMN_NAME, 'FM000') FROM TABLE_NAME;
Both is working fine. but don't know which one would be the best choice.