I have below query which gives me output like -$97.90 or $11.00
Select TRIM(to_char(pen_amt,'$999,999,999,999,999.99')) as PenAmount
from transact;
Now, if the value starts with -, I want to remove - and encapsulate the same as ($97.90). How can I do this?
Also using the PR format model element, which surrounds positive (and zero) values with single spaces and negative numbers with angled brackets: You can follow that up with the TRANSLATE function if you really need the format you requested. (Notice that if I use TRANSLATE, I don't need to use TRIM, or the format model modifier FM in TO_CHAR, which would do the same thing; I can simply remove the spaces with TRANSLATE.)
with
test_data (amount) as (
select 320.88 from dual union all
select -309 from dual
)
select amount,
to_char(amount, '$999,999,999.99pr') as pr_formatted_amount,
translate(to_char(amount, '$999,999,999.99pr'), '<> ', '()')
as my_formatted_amount
from test_data
;
AMOUNT PR_FORMATTED_AMOUNT MY_FORMATTED_AMOUNT
---------- ------------------- -------------------
320.88 $320.88 $320.88
-309 <$309.00> ($309.00)
If you can use <> brackets then PR will do the trick.
Select TRIM(to_char(pen_amt,'$999,999,999,999,999.99PR')) as PenAmount from transact;
Use a CASE statement to check if the value starts with -:
Select
case when TRIM(to_char(pen_amt,'$999,999,999,999,999.99')) like '-%'
then '(' || substr(TRIM(to_char(pen_amt,'$999,999,999,999,999.99')), 2) || ')'
else TRIM(to_char(pen_amt,'$999,999,999,999,999.99'))
end as PenAmount
from transact;
Or by directly checking if pen_amt is negative:
Select
case when pen_amt < 0
then '(' || TRIM(to_char(abs(pen_amt),'$999,999,999,999,999.99')) || ')'
else TRIM(to_char(pen_amt,'$999,999,999,999,999.99'))
end as PenAmount
from transact;
Use a combination of PR in the mask along with TRANSLATE to convert angle brackets to parentheses.
Select translate(to_char(-12345.6789,'FM$999,999,999,999,999.99PR'),'<>','()') as PenAmount from dual;
Select translate(to_char(23456.789,'FM$999,999,999,999,999.99PR'),'<>','()') as PenAmount from dual;
($12,345.68)
$23,456.79
The PR mask puts angle brackets for negative numbers.
The FM mask tells Oracle to show the minimal text (i.e. TRIM it).
Related
I have a table, inside there is a field called x, the x field contain value of '1+2', '1+3' etc, how to get these value and calculate it and save into another field?
For simple arithmetic expressions - and depending on your Oracle version - you could use xmlquery to evaluate. Note that / has special meaning in xml, the operator for division is the keyword div - so you need a replace in case you may have forward slashes in the arithmetic expression. (If you don't have any divisions, you can simplify the query by removing the call to replace.)
Here is an example - including the test data at the top, in a with clause (not part of the solution!)
with
test_data (str) as (
select '1 + 3' from dual union all
select '3 * 5 - 2' from dual union all
select '2/4*6' from dual union all
select '3 * (1 - 3)' from dual
)
select str, xmlquery(replace(str, '/', ' div ') returning content).getNumberVal()
as evaluated_expression
from test_data;
STR EVALUATED_EXPRESSION
----------- --------------------
1 + 3 4
3 * 5 - 2 13
2/4*6 3
3 * (1 - 3) -6
If you only have valid PL/SQL arithmetic expressions in your formulas, then you can use EXECUTE IMMEDIATE to evaluate them.
For this, you'll need to create a function:
create or replace function eval_expression(p_expression in varchar2)
return number is
query varchar2(100);
result number;
begin
query := 'select ' || p_expression || ' from dual';
execute immediate query
into result;
return result;
end eval_expression;
Then you can use this function in UPDATE query:
update t
--val is another field
set t.val = eval_expression(t.x)
Naturally, with EXECUTE IMMEDIATE this query won't be extremely efficient, but it'll work. Also, with dynamic queries we're going into unsafe territory, so make sure that you don't have malicious code among your formulas.
Also, see "Evaluate Expression" on Ask TOM. Tom Kyte used a slightly more civilized approach (dbms_sql package) and created a package with a single variable support.
If the case you have mentioned needs to be considered then I will suggest you use the following query:
select
xmlquery('3+4'
returning content
).getNumberVal()
from
dual;
If More operators are involved except division then the aforementioned query will work but if the division operator is also involved then you must have to replace "/" with " div " keyword. Something like the following:
select
xmlquery(
replace( '20/5', '/', ' div ')
returning content
).getNumberVal()
from
dual;
Now, you can use it in your update statement or anywhere else.
update tab
set tab.result_column_name = xmlquery(t.your_expr_column_name
returning content
).getNumberVal()
update tab
set tab.result_column_name = xmlquery(
replace( t.your_expr_column_name, '/', ' div ')
returning content
).getNumberVal()
Demo
Cheers!!
I need help
i have records 123,456,789 in rows when i am execute like
this one is working
select * from table1 where num1 in('123','456')
but when i am execute
select * from table1 where num1 in(select value from table2)
no resultset found - why?
Check the DataType varchare2 or Number
try
select * from table1 where num1 in(select to_char(value) from table2)
Storing comma separated values could be the cause of problem.
You can try using regexp_substr to split comma.
First and foremost, an important thing to remember: Do not store numbers in character datatypes. Use NUMBER or INTEGER. Secondly, always prefer VARCHAR2 datatype over CHAR if you wish to store characters > 1.
You said in one of your comments that num1 column is of type char(4). The problem with CHAR datatype is that If your string is 3 characters wide, it stores the record by adding extra 1 space character to make it 4 characters. VARCHAR2 only stores as many characters as you pass while inserting/updating and are not blank padded.
To verify that you may run select length(any_char_col) from t;
Coming to your problem, the IN condition is never satisfied because what's actually being compared is
WHERE 'abc ' = 'abc' - Note the extra space in left side operator.
To fix this, one good option is to pad the right side expression with as many spaces as required to do the right comparison.The function RPAD( string1, padded_length [, pad_string] ) could be used for this purpose.So, your query should look something like this.
select * from table1 where num1 IN (select rpad(value,4) from table2);
This will likely utilise an index on the column num1 if it exists.
The other one is to use RTRIM on LHS, which is only useful if there's a function based index on RTRIM(num1)
select * from table1 where RTRIM(num1) in(select value from table2);
So, the takeaway from all these examples is always use NUMBER types to store numbers and prefer VARCHAR2 over CHAR for strings.
See Demo to fully understand what's happening.
EDIT : It seems You are storing comma separated numbers.You could do something like this.
SELECT *
FROM table1 t1
WHERE EXISTS (
SELECT 1
FROM table2 t2
WHERE ',' ||t2.value|| ',' LIKE '%,' || rtrim(t1.num1) || ',%'
);
See Demo2
Storing comma separated values are bound to cause problems, better change it.
Let me tell you first,
You have stored values in table2 which is comma seperated.
So, how could you match your data with table1 and table2.
Its not Possible.
That's why you did not get any values in result set.
I found the Solution using string array
SELECT T.* FROM TABLE1 T,
(SELECT TRIM(VALUE)AS VAL FROM TABLE2)TABLE2
WHERE
TRIM(NUM1) IN (SELECT COLUMN_VALUE FROM TABLE(FUNC_GETSTRING_ARRAY(TABLE2.VAL)))
thanks
How can I find if the fifth position is a letter and thus not a number using Oracle ?
My last try was using the following statement:
REGEXP_LIKE (table_column, '([abcdefghijklmnopqrstuvxyz])');
Perhaps you'd rather check whether 5th position contains a number (which means that it is not something else), i.e. do the opposite of what you're doing now.
Why? Because a "letter" isn't only ASCII; have a look at the 4th row in my example - it contains Croatian characters and these aren't between [a-z] (nor [A-Z]).
SQL> with test (col) as
2 (select 'abc_3def' from dual union all
3 select 'A435D887' from dual union all
4 select '!#$%&/()' from dual union all
5 select 'ASDĐŠŽĆČ' from dual
6 )
7 select col,
8 case when regexp_like(substr(col, 5, 1), '\d+') then 'number'
9 else 'not a number'
10 end result
11 from test;
COL RESULT
------------- ------------
abc_3def number
A435D887 not a number
!#$%&/() not a number
ASDĐŠŽĆČ not a number
SQL>
Anchor to the start of the string else you may get unexpected results. This works, but remove the caret (start of string anchor) and it returns 'TRUE'! Note it uses the case-insensitive flag of 'i'.
select 'TRUE'
from dual
where regexp_like('abcd4fg', '^.{4}[A-Z]', 'i');
Yet another way to do it:
regexp_like(table_column, '^....[[:alpha:]]')
Using the character class [[:alpha:]] will pick up all letters upper case, lower case, accented and etc. but will ignore numbers, punctuation and white space characters.
If what you care about is that the character is not a number, then use
not regexp_like(table_column, '^....[[:digit:]]')
or
not regexp_like(table_column, '^....\d')
Try:
REGEXP_LIKE (table_column, '^....[a-z]')
Or:
SUBSTR (table_column, 5, 1 ) BETWEEN 'a' AND 'z'
i have a string 'MCDONALD_YYYYMMDD.TXT' i need to use regular expressions and append the '**' after the letter 'D' in the string given . (i.e In the string at postion 9 i need to append '*' based on a column value 'star_len'
if the star_len = 2 the o/p = ''MCDONALD??_YYYYMMDD.TXT'
if the star_len = 1 the o/p = ''MCDONALD?_YYYYMMDD.TXT'
with
inputs ( filename, position, symbol, len ) as (
select 'MCDONALD_20170812.TXT', 9, '*', 2 from dual
)
-- End of simulated inputs (for testing purposes only, not part of the solution).
-- SQL query begins BELOW THIS LINE.
select substr(filename, 1, position - 1) || rpad(symbol, len, symbol)
|| substr(filename, position) as new_str
from inputs
;
NEW_STR
-----------------------
MCDONALD**_20170812.TXT
select regexp_replace('MCDONALD_YYYYMMDD.TXT','MCDONALD','MCDONALD' ||
decode(star_len,1,'*',2,'**'))
from dual
This is how you could do it. I don't think you need it as a regular expression though if it is always going to be "MCDONALD".
EDIT: If you need to be providing the position in the string as well, I think a regular old substring should work.
select substr('MCDONALD_YYYYMMDD.TXT',1,position-1) ||
decode(star_len,1,'*',2,'**') || substr('MCDONALD_YYYYMMDD.TXT',position)
from dual
Where position and star_len are both columns in some table you provide(instead of dual).
EDIT2: Just to be more clear, here is another example using a with clause so that it runs without adding a table in.
with testing as
(select 'MCDONALD_YYYYMMDD.TXT' filename,
9 positionnum,
2 star_len
from dual)
select substr(filename,1,positionnum-1) ||
decode(star_len,1,'*',2,'**') ||
substr(filename,positionnum)
from testing
For the fun of it, here's a regex_replace solution. I went with a star since that what your variable was called even though your example used a question mark. The regex captures the filename string in 2 parts, the first being from the start up to 1 character before the position value, the second the rest of the string. The replace puts the captured parts back together with the stars in between.
with tbl(filename, position, star_len ) as (
select 'MCDONALD_20170812.TXT', 9, 2 from dual
)
select regexp_replace(filename,
'^(.{'||(position-1)||'})(.*)$', '\1'||rpad('*', star_len, '*')||'\2') as fixed
from tbl;
I would like to get:
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
from the following expression
LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Does someone know how I can get this using regexp_substr ?
EDIT
Basically I have a field which has 7 sets each separated by _ . The string I gave is just one example. I wanted to retrieve everything after the second _ . There is no fixed character length so I can not use a substr function. Hence I was using regexp_substr. I was able to get away by using a simplified version
Select FILE_NAME, ( (REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,3)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,4)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,5)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,6)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+',1,7)) ) as RegExp
from tbl
Here is some more data from the FILE_NAME field
LAST_FIRST_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
SMITH_JOHN_82961_0130BPQX9QZN9G4P5RDTPA9HR4R_MR_1_1of1
LASTNAME_FIRSTNAME_99999_01V0MU4XUQK0Y24Y9RYTFA7W1CM_MR_3_1of1
To get everything after the second underscore, you do not need regular expressions, but can use something like the following:
select substr(FILE_NAME, instr(FILE_NAME, '_', 1, 2) +1 ) from tbl
The instr returns the position of the second occurrence of '_', starting by the first character; the substr simply gets everything starting from the position given by instr + 1
From your Requirement, you can just go ahead and use the simple SUBSTRfunction. Its faster, and it addresses the simple need to remove the String LASTNAME_FIRSTNAME.
select substr('LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1', 20) data_string
from dual;
Output:
data_string
-----------------
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Unless you have another underlying logic you need to address?
Kindly clarify so i can edit the answer accordingly.