I need to give format to the following number:
1234567.89
as
1 234 567.89
I already tried:
select regexp_replace( '1234567.89', '(...)', '\1 ' ) from dual;
But its starting from left to right the counting and it's ignoring the decimal dot.
Thanks in advance.
Best regards.
SELECT TO_CHAR(10000,'99G999D99MI',
'NLS_NUMERIC_CHARACTERS = ''. ''
NLS_CURRENCY = '' ') "Amount"
FROM DUAL;
you can do it like this
select replace(to_char(1234567.89, '9,999,999,999,999,999.99'), ',', ' ') x from dual
Only in this case, you need to know how big is the biggest number you will have. If lets say, your biggest number is in millions, make format model in billions, so it covers either exact number of digits or more
Related
I have a column in a table ident_nums that contains different types of ids. I need to remove special characters(e.g. [.,/#&$-]) from that column and replace them with space; however, if the special characters are found at the beginning of the string, I need to remove it without placing a space. I tried to do it in steps; first, I removed the special characters and replaced them with space (I used
REGEXP_REPLACE) then found the records that contain spaces at the beginning of the string and tried to use the TRIM function to remove the white space, but for some reason is not working that.
Here is what I have done
Select regexp_replace(id_num, '[:(),./#*&-]', ' ') from ident_nums
This part works for me, I remove all the unwanted characters from the column, however, if the string in the column starts with a character I don't want to have space in there, I would like to remove just the character, so I tried to use the built-in function TRIM.
update ident_nums
set id_num = TRIM(id_num)
I'm getting an error ORA-01407: can't update ident_nums.id_num to NULL
Any ideas what I am doing wrong here?
It does work if I add a where clause,
update ident_nums
set id_num = TRIM(id_num) where id = 123;
but I need to update all the rows with the white space at the beginning of the string.
Any suggestions are welcome.
Or if it can be done better.
The table has millions of records.
Thank you
Regexp can be slow sometimes so if you can do it by using built-in functions - consider it.
As #Abra suggested TRIM and TRANSLATE is a good choice, but maybe you would prefer LTRIM - removes only leading spaces from string (TRIM removes both - leading and trailing character ). If you want to remove "space" you can ommit defining the trim character parameter, space is default.
select
ltrim(translate('#kdjdj:', '[:(),./#*&-]', ' '))
from dual;
select
ltrim(translate(orginal_string, 'special_characters_to_remove', ' '))
from dual;
Combination of Oracle built-in functions TRANSLATE and TRIM worked for me.
select trim(' ' from translate('#$one,$2-zero...', '#$,-.',' ')) as RESULT
from DUAL
Refer to this dbfiddle
I think trim() is the key, but if you want to keep only alpha numerics, digits, and spaces, then:
select trim(' ' from regexp_replace(col, '[^a-zA-Z0-9 ]', ' ', 1, 0))
regexp_replace() makes it possible to specify only the characters you want to keep, which could be convenient.
Thanks, everyone, It this query worked for me
update update ident_nums
set id_num = LTRIM(REGEXP_REPLACE(id_num, '[:space:]+', ' ')
where REGEXP_LIKE(id_num, '^[ ?]')
this should work for you.
SELECT id_num, length(id_num) length_old, NEW_ID_NUM, length(NEW_ID_NUM) len_NEW_ID_NUM, ltrim(NEW_ID_NUM), length(ltrim(NEW_ID_NUM)) length_after_ltrim
FROM (
SELECT id_num, regexp_replace(id_num, '[:(),./#*&-#]', ' ') NEW_ID_NUM FROM
(
SELECT '1234$%45' as id_num from dual UNION
SELECT '#SHARMA' as id_num from dual UNION
SELECT 'JACK TEST' as id_num from dual UNION
SELECT 'XYZ#$' as id_num from dual UNION
SELECT '#ABCDE()' as id_num from dual -- THe 1st character is space
)
)
Struggle to design a regular expression to filter field value from varchar2 to number, so that it can remove all non-digit and only left the last period in the string, so that
"about 1,000.00" return 1000.00 or 1000
"3,000,000.000" return 300000.000 or 3000000
"3.000.000.000" return return 3000000.000 or 3000000
"a^*3^%*(C4.5d*9" return 34.59
Any method just change the string into accurate convertible string that can be converted by to_number()
I use
SELECT REGEXP_REPLACE(field_value, '[^0-9\.]+', '') from dual;
but can't resolve the 3rd case....
Because the regex in oracle are somewhat limited I don't think it's possible only using regexp_replace. You could do a workaround like this:
SELECT
CASE
WHEN last_dot < 2 THEN digits_and_dots
ELSE REPLACE(SUBSTR(digits_and_dots, 1, last_dot - 1), '.') ||
SUBSTR(digits_and_dots, last_dot)
END
FROM (
SELECT
INSTR(digits_and_dots, '.', -1) last_dot,
digits_and_dots
FROM (
SELECT
REGEXP_REPLACE(field_value, '[^0-9\.]+', '') digits_and_dots
FROM DUAL
) t
) o
Here's a way to do it, assuming there is one decimal character. The value you are working with is a string so I think of the decimal that we want to keep as a separator of the string and split it into 2 parts based on that. The first part is all characters leading up to but not including the last decimal, the second part is the last decimal and all characters after it. Then apply the replace, getting rid of everything that is not a number from the first part, and everything that is not a number or a decimal from the second part, then concatenate them together. Needs more testing with varied inputs but you get the idea. All these regular expressions are kind of expensive though so I doubt this will be the fastest solution.
with tbl(str) as (
select 'about 1,000.00' from dual union
select '3,000,000.000' from dual union
select '3.000.000.000' from dual union
select 'a^*3^%*(C4.5d*9' from dual
)
select str original,
regexp_replace(regexp_substr(str, '^(.*)\.', 1, 1, NULL, 1), '[^0-9]+', '') ||
regexp_replace(regexp_substr(str, '.*(\..*)$', 1, 1, NULL, 1), '[^0-9\.]+', '') Converted
from tbl;
SQL> /
ORIGINAL CONVERTED
--------------- ---------------
3,000,000.000 3000000.000
3.000.000.000 3000000.000
a^*3^%*(C4.5d*9 34.59
about 1,000.00 1000.00
SQL>
Shortest way is as follows:
select regexp_substr('a^*3^%*(C4.5d*9s','\d+\.\d+') from dual;
or
select regexp_replace('a^*3^%*(C4.5d*9s', '[^0.0-9]', '') from dual;
I have the below view:
CREATE OR REPLACE VIEW viewA ("col1", "col2") AS
SELECT DISTINCT CAST("col1" AS CHAR(1)),
CAST(to_char("col2",'00.0000') AS char(7))
FROM tableA
the col2 has data like 22.33 or 2.3 or 0.2345 or 2 but, four digits in dec and 2 digits in number.
It has to be written into a file with fixed length of 7 digits including decimal. Hence i wrote col2, '00.0000', but the number format'23.234' is written into col2 as 23.234 without any trailing zero.
Your format code of 00.0000 should include the fourth decimal place for 23.234; it's always worked for me. I'm using Oracle 11.
The problem I got when I tried doing CAST(TO_CHAR(23.234, '00.0000') AS CHAR(7)) was the error ORA-25137: Data value out of range. This happens because because the TO_CHAR returns a string of length 8:
SQL> SELECT '[' || TO_CHAR(23.234, '00.0000') || ']' FROM DUAL
'['||TO_CH
----------
[ 23.2340]
TO_CHAR leaves a space at the beginning in case the number is negative, in which case it will put a minus sign there. You can get rid of the leading space by using the FM modifier in the format string:
SQL> SELECT '[' || TO_CHAR(23.234, 'FM00.0000') || ']' FROM DUAL
'['||TO_CH
----------
[23.2340]
This is all a long way of saying "try this instead" - the only change is the FM in the TO_CHAR format string:
CREATE OR REPLACE VIEW viewA ("col1", "col2") AS
SELECT DISTINCT
CAST("col1" AS CHAR(1)),
CAST(to_char("col2",'FM00.0000') AS char(7))
FROM tableA
One final note: enclosing the column names with double quotes makes them case sensitive, and that often leads to trouble. I'd recommend losing the double quotes if you can.
You need to use the RPAD function that would add trailing zeros for you
CREATE OR REPLACE VIEW viewA ("col1", "col2") AS
SELECT DISTINCT CAST("col1" AS CHAR(1)),
RPAD(CAST(to_char("col2",'00.0000') AS char(7)),7,'0')
FROM tableA
But you might face a problem if your number did not have a decimal value, for example assuming the value is 12 you will end up with 1200000 but maybe this would give you an idea
Given data in a column which look like this:
00001 00
00026 00
I need to use SQL to remove anything after the space and all leading zeros from the values so that the final output will be:
1
26
How can I best do this?
Btw I'm using DB2
This was tested on DB2 for Linux/Unix/Windows and z/OS.
You can use the LOCATE() function in DB2 to find the character position of the first space in a string, and then send that to SUBSTR() as the end location (minus one) to get only the first number of the string. Casting to INT will get rid of the leading zeros, but if you need it in string form, you can CAST again to CHAR.
SELECT CAST(SUBSTR(col, 1, LOCATE(' ', col) - 1) AS INT)
FROM tab
In DB2 (Express-C 9.7.5) you can use the SQL standard TRIM() function:
db2 => CREATE TABLE tbl (vc VARCHAR(64))
DB20000I The SQL command completed successfully.
db2 => INSERT INTO tbl (vc) VALUES ('00001 00'), ('00026 00')
DB20000I The SQL command completed successfully.
db2 => SELECT TRIM(TRIM('0' FROM vc)) AS trimmed FROM tbl
TRIMMED
----------------------------------------------------------------
1
26
2 record(s) selected.
The inner TRIM() removes leading and trailing zero characters, while the outer trim removes spaces.
This worked for me on the AS400 DB2.
The "L" stands for Leading.
You can also use "T" for Trailing.
I am assuming the field type is currently VARCHAR, do you need to store things other than INTs?
If the field type was INT, they would be removed automatically.
Alternatively, to select the values:
SELECT (CAST(CAST Col1 AS int) AS varchar) AS Col1
I found this thread for some reason and find it odd that no one actually answered the question. It seems that the goal is to return a left adjusted field:
SELECT
TRIM(L '0' FROM SUBSTR(trim(col) || ' ',1,LOCATE(' ',trim(col) || ' ') - 1))
FROM tab
One option is implicit casting: SELECT SUBSTR(column, 1, 5) + 0 AS column_as_number ...
That assumes that the structure is nnnnn nn, ie exactly 5 characters, a space and two more characters.
Explicit casting, ie SUBSTR(column,1,5)::INT is also a possibility, but exact syntax depends on the RDBMS in question.
Use the following to achieve this when the space location is variable, or even when it's fixed and you want to make a more robust query (in case it moves later):
SELECT CAST(SUBSTR(LTRIM('00123 45'), 1, CASE WHEN LOCATE(' ', LTRIM('00123 45')) <= 1 THEN LEN('00123 45') ELSE LOCATE(' ', LTRIM('00123 45')) - 1 END) AS BIGINT)
If you know the column will always contain a blank space after the start:
SELECT CAST(LOCATE(LTRIM('00123 45'), 1, LOCATE(' ', LTRIM('00123 45')) - 1) AS BIGINT)
both of these result in:
123
so your query would
SELECT CAST(SUBSTR(LTRIM(myCol1), 1, CASE WHEN LOCATE(' ', LTRIM(myCol1)) <= 1 THEN LEN(myCol1) ELSE LOCATE(' ', LTRIM(myCol1)) - 1 END) AS BIGINT)
FROM myTable1
This removes any content after the first space character (ignoring leading spaces), and then converts the remainder to a 64bit integer which will then remove all leading zeroes.
If you want to keep all the numbers and just remove the leading zeroes and any spaces you can use:
SELECT CAST(REPLACE('00123 45', ' ', '') AS BIGINT)
While my answer might seem quite verbose compared to simply SELECT CAST(SUBSTR(myCol1, 1, 5) AS BIGINT) FROM myTable1 but it allows for the space character to not always be there, situations where the myCol1 value is not of the form nnnnn nn if the string is nn nn then the convert to int will fail.
Remember to be careful if you use the TRIM function to remove the leading zeroes, and actually in all situations you will need to test your code with data like 00120 00 and see if it returns 12 instead of the correct value of 120.
Does anyone know how to turn this string: "Smith, John R"
Into this string: "jsmith" ?
I need to lowercase everything with lower()
Find where the comma is and track it's integer location value
Get the first character after that comma and put it in front of the string
Then get the entire last name and stick it after the first initial.
Sidenote - instr() function is not compatible with my version
Thanks for any help!
Start by writing your own INSTR function - call it my_instr for example. It will start at char 1 and loop until it finds a ','.
Then use as you would INSTR.
The best way to do this is using Oracle Regular Expressions feature, like this:
SELECT LOWER(regexp_replace('Smith, John R',
'(.+)(, )([A-Z])(.+)',
'\3\1', 1, 1))
FROM DUAL;
That says, 1) when you find the pattern of any set of characters, followed by ", ", followed by an uppercase character, followed by any remaining characters, take the third element (initial of first name) and append the last name. Then make everything lowercase.
Your side note: "instr() function is not compatible with my version" doesn't make sense to me, as that function's been around for ages. Check your version, because Regular Expressions was only added to Oracle in version 9i.
Thanks for the points.
-- Stew
instr() is not compatible with your version of what? Oracle? Are you using version 4 or something?
There is no need to create your own function, and quite frankly, it seems a waste of time when this can be done fairly easily with sql functions that already exist. Care must be taken to account for sloppy data entry.
Here is another way to accomplish your stated goal:
with name_list as
(select ' Parisi, Kenneth R' name from dual)
select name
-- There may be a space after the comma. This will strip an arbitrary
-- amount of whitespace from the first name, so we can easily extract
-- the first initial.
, substr(trim(substr(name, instr(name, ',') + 1)), 1, 1) AS first_init
-- a simple substring function, from the first character until the
-- last character before the comma.
, substr(trim(name), 1, instr(trim(name), ',') - 1) AS last_name
-- put together what we have done above to create the output field
, lower(substr(trim(substr(name, instr(name, ',') + 1)), 1, 1)) ||
lower(substr(trim(name), 1, instr(trim(name), ',') - 1)) AS init_plus_last
from name_list;
HTH,
Gabe
I have a hard time believing you don’t have access to a proper instr() but if that’s the case, implement your own version.
Assuming you have that straightened out:
select
substr(
lower( 'Smith, John R' )
, instr( 'Smith, John R', ',' ) + 2
, 1
) || -- first_initial
substr(
lower( 'Smith, John R' )
, 1
, instr( 'Smith, John R', ',' ) - 1
) -- last_name
from dual;
Also, be careful about your assumption that all names will be in that format. Watch out for something other than a single space after the comma, last names having data like “Parisi, Jr.”, etc.