I have this number 0101110 but i need only 5 digits. for exemple (01110)
i used this
select to_char(trim('0101110')::int,'99999V')
but PG returns #####
if someone can help me , thanks a lot.
How do you have a number that starts with a zero? I suspect your "number" is really being stored as a string.
If so, just use right() to take the rightmost five characters:
select right(col, 5)
If the value is really stored as a number, then convert it to a string first.
Related
Here's what i'm having trouble with. I have a table where i have the column weight, the data in this column is varchar2, example: 60 kg.
When i try to make a WHERE condition asking to retrieve everything below 60 kg, it doesn't give me the wanted results.
I tried to convert using to_number but it's not working and giving me error "Invalid Number". I also tried to cut the kg with SUBSTR and convert, and doesn't work as well.
Any suggestions?
Thanks in advance.
You can extract leading digits and convert that to a number:
select to_number(regexp_substr(weight, '^[0-9]*'))
from t;
The regular expression is starting from the beginning of the string (the ^) and extracting digits. The * is for any number of digits. Regular expression match is "greedy" by default, meaning that it will keep matching as many digits in a row as there are.
If you have decimal places or negative values, this might be a tad bit more complicated.
I have column in a postgresql database. They are lottery numbers. Four digits in length to be exact. Initially I had the datatype of the column as int. I inserted all the lottery numbers. After I inserted all the numbers I realize it chopped off my zeros. For instance 0925 is 925. I fixed the datatype to be varchar but now I need to figure out how to fix it from int to varchar with the same data. The data needs to be 4 digits in length. I was trying to just figure out how many problem numbers there are and I couldn't write a select statement that told me how many rows have less than 4 digits.
How should I go about this?
Thanks.
I was trying to just figure out how many problem numbers there are and I couldn't write a select statement that told me how many rows have less than 4 digits.
SELECT COUNT(*)
FROM lottery
WHERE char_length(x) < 4
See it working online: sqlfiddle
To fix them, you may find lpad useful. Note that the WHERE clause is not actually needed.
UPDATE lottery
SET x = lpad(x, 4, '0')
See it working online: sqlfiddle
Format your numbers with to_char():
SELECT to_char(123, 'FM0000');
You might even just leave them as integer and use the expression in queries.
Or, to convert your column back from integer to text in place:
ALTER TABLE tbl ALTER column col TYPE text USING to_char(col, 'FM0000');
Since you seem to have already converted the numbers to varchar, the expression needs an additional cast to integer:
SELECT to_char(col::int, 'FM0000')
FROM tbl;
I have a question that bothers me. How can i convert a varchar to number when inside the varchar value consists of alphabets.
For my varchar price column values:
14dollars10cents
15dollars20cents
By converting it to varchar to number price column, the values should be:
1410
1520
I know that if the varchar does not consists any alphabets, it can auto convert by"
SELECT CONVERT(INT, PRICE) FROM Table
Is there any way to get rid of the alphabets in the middle as I would like to do mathematical function on it.
Updated attempt of putting fixed point number in:
SELECT CAST (Replace(REPLACE(PRICE, 'dollars', '.'),'cents','') AS Number(4,2)))
FROM TEST;
Thanks
You could just use REGEXP_REPLACE to remove all non digit characters:
SELECT REGEXP_REPLACE(price, '[^[:digit:]]')
FROM table;
To then convert this to a number:
SELECT TO_NUMBER(REGEXP_REPLACE(price, '[^[:digit:]]'))
FROM table;
If you want to add the point in then you can do that with REGEXP_REPLACE too:
SELECT TO_NUMBER(REGEXP_REPLACE(val, '^([0-9]*)([^[:digit:]]*)([0-9]*)(.*)$', '\1.\3'))
FROM table;
Voila...
SELECT CAST(REPLACE(YourVarcharCol, 'dollars', '') AS INT) FROM Table
The issue with this is it will break if the varchar still contains alpha-numeric characters.
How about using translate to strip out the unwanted characters.
SELECT TO_NUMBER(TRANSLATE('14dollars10cents','1234567890dolarscents','1234567890')) FROM DUAL
No I don't think there is direct way.
you can do string parsing to get your integer value.
i could not understand why following code
SQL>
Select to_number('1234.64', '9999.9') from Dual;
returns this number 1234.6?is it something like rounding ,truncation or?please help me to understand this code,i know to_number functions,i have used many times this code for simple chars,but here it is not clear anything
This looks a lot like Oracle, but I suspect that the result would be similar in any SQL that used to_number.
The to_number function takes two arguments: the string to be converted to a number, and the format string for the conversion.
In the example, '12345.64' is the string to be converted, while '9999.9' is the format string. In this format string, a 9 stands for a digit while a . stands for the decimal point.
So the function is asking to convert the string '12345.64' to a number with up to 4 digits to the right of the decimal point, and only 1 digit after the decimal point.
The second argument is optional - under normal circumstances, I would omit it.
You should use
SELECT to_number('1234.64', '9999.99') from Dual;
Your mask tells engine you want just one decimal, so number gets rounded.
If you want to get exact number, don't specify any mask:
SELECT to_number('1234.64') from Dual;
I have a column in my table which showing an amount. The amount is varying from one column to another and they are more than 15 digits.
What is the best way to format the number to show commas and decimal points?
My query is
select amount from ccamounts
How can I format the number
205511892078
to show as
205,511,892,078
and if there is a radix point it will also appear.
I believe you can use TO_CHAR to do this, the issue is that this is just a formatting function within SQL. It requires that your number is always going to be in the same format.
taking the example above you could do
TO_CHAR('205511892078', '999,999,999,999')
and this would format the number as you have specified, with a decimal place this can be done aswell but the decimal needs to be specified:
TO_CHAR('20551189207842', '999,999,999,999.99')
which would give you 205,511,892,078.42
I think if the field length is going to vary sql will just ignore anything that doesn't fit into the format string (It's a mask). Perhaps you want to consider formatting the number in this case on whichever front end you may be using?
I would format the number in the UI / Reporting tool / Presentation layer not Oracle
but if you MUST format it in oracle try:
SELECT
CASE WHEN INSTR( TO_CHAR(205511892078),'.')>0 THEN
TO_CHAR(205511892078 ,'999,999,999,999.99')
ELSE
TO_CHAR(205511892078 ,'999,999,999,999')
END
FROM DUAL
this will return the number as a string.
declare #d3 decimal (10, 2)
set #d3 = 12309809.5494
SELECT convert(varchar(15),cast(CAST(ROUND(#d3,2,1) AS DECIMAL (30,2)) as money),1) as Value
SELECT CAST(ROUND(convert(varchar(30), cast(#d3 as money),2),2,1) AS DECIMAL (30,2)) as Value
Output:
12,309,809.55
12309809.55