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'
Related
I need to display different results in the next format, example:
40000000 to 40'000,000
I tried using this, but when i try 2 differents group separators i get the "invalid number format model" error:
select to_char(9999999999, '9g999g99999g9', 'NLS_NUMERIC_CHARACTERS='',.''')
from dual;
Also tried using substr and replace but it doesnt work in all the cases (like when the result is 3000000 or 700000000).
This works but it is not the optimal solution.:
SELECT substr(replace('40,000,000',',',''''),0,length(40000000)-2)|| substr('40,000,000',-4) from dual;
What the actual select look like if i use the previous code.
SELECT substr(replace(to_char(oTOTAL_SENIOR, '999,999,999'),',',''''),0,length(oTOTAL_SENIOR)-2)|| substr(to_char(oTOTAL_SENIOR, '999,999,999'),-4) from dual
The previous select gets bugged when i use substr replace and to_char together because of the '999,999,999'.
I also tried using regexp_replace but im not good at it.
I know i need to replace everything but the last 4 characters (,000) but i dont know how.
Any help will be aprreciated.
You'll need someone smarter than me to do that properly, but - meanwhile, see whether this helps.
I'm on XE 11g; in order to avoid "no more data to read from socket" error I got during the final steps of this query (I believe it was due to two string reversings), I created my own "reverse" function. There's undocumented reverse function, but I don't want to use it.
Basically, it reverses a string you pass as a parameter. What do I need it for? I found it simpler to reverse values, split them to 3-by-3-by-3 characters and apply those "strange" separators you want. Also, it makes the whole code simpler. It can be done without it, but - as I said - no more data to read from socket won't allow it. Sorry about that.
Now, someone will say: why didn't you (meaning: me, LF) do that using PL/SQL completely and put everything into a function? No particular reason & no problem in doing it, if necessary.
OK, here it is:
SQL> create or replace function f_reverse (par_string in varchar2)
2 return varchar2
3 is
4 retval varchar2(20);
5 begin
6 select listagg(substr(par_string, level, 1))
7 within group (order by level desc)
8 into retval
9 from dual
10 connect by level <= length(par_string);
11
12 return retval;
13 end;
14 /
Function created.
SQL> select f_reverse('1234') from dual;
F_REVERSE('1234')
---------------------------------------------------------------------
4321
SQL>
Finally, this is what you want:
SQL> with test (id, col) as
2 (select 1, 40100200 from dual union all
3 select 2, 2300400 from dual union all
4 select 3, 700500 from dual union all
5 select 4, 25700 from dual union all
6 select 5, 6300 from dual union all
7 select 6, 555 from dual
8 )
9 select id,
10 regexp_replace(
11 f_reverse(
12 substr(f_reverse(col), 1, 3) ||','||
13 substr(f_reverse(col), 4, 3) || chr(39) ||
14 substr(f_reverse(col), 7)
15 ), '^[^0-9]+', '') result
16 from test;
ID RESULT
---------- ----------
1 40'100,200
2 2'300,400
3 700,500
4 25,700
5 6,300
6 555
6 rows selected.
SQL>
What does it do?
lines #1 - 7 - sample data
lines #9 onward is useful code
lines #12 - 14 - splitting reversed sample data into substrings 3 characters in length
line #12 - concatenated , as this is thousands separator
line #13 - concatenaded chr(13) which is ', a millions separator
line #11 - reversing concatenated "reversed" string back
line #10 - removing possible non-numeric characters from the beginning of the result (those are separators for values that are shorter than "thousand" or "million")
In APEX 5.1 looking for validation script help to only allow alphabetic characters and a space or hyphen for a person's name. (i.e. no numbers or special characters other than a possible dash allowed).
I have a working validation for another field which allows alphanumeric but not special characters! Validation type is "Item does NOT contain any of the characters in Value". In the value field the following is placed... !##$%^&*()-_+=[]{};:'",<.>/?\|
I'd suggest you to use validation whose type is a function that returns Boolean and looks like this:
return regexp_like(:P1_ITEM_NAME, '^[A-Za-z -]+$');
What does it do?
^ is anchor to the beginning of the string
A-Z accepts capital letters
a-z accepts lowercase letters
space is ... well, a space
- is ... well, a hyphen
+ - repeat those characters many times
$ is anchor to the end of the string
For example:
SQL> with test (item) as
2 (select 'aBCd' from dual union -- ok
3 select 'little foot' from dual union -- ok
4 select 'reb-ecca' from dual union -- ok
5 select 'lit123foot' from dual union -- wrong
6 select 'abc$3' from dual union -- wrong
7 select 'xy.)z' from dual union -- wrong
8 select '123-234' from dual -- wrong
9 )
10 select item
11 from test
12 where regexp_like(item, '^[A-Za-z -]+$');
ITEM
-----------
aBCd
little foot
reb-ecca
SQL>
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;
how to display second or third letter in each word of sentence as CAPITAL letter in Oracle.
Testdata:
hyderabad
Output:
hYderabad or hyDerabad
To make 2nd character to upper case
select SUBSTR(test_data,1,1)||INITCAP(SUBSTR(test_data,2)) from test_table;
To make the 3rd character to upper case, increment the last argument of both SUBSTR with 1.
Example
select SUBSTR(test_data,1,2)||INITCAP(SUBSTR(test_data,3)) from test_table;
Use this for 2nd digit as upper case.
WITH TBL(TESTDATA) AS
( SELECT 'hyderabad' FROM DUAL UNION
SELECT 'pune' FROM DUAL UNION
SELECT '223jjjj22' FROM DUAL
)
SELECT substr(testdata,1,1)||upper(substr(testdata,2,1))
||substr(testdata,3,length(testdata)-2)
as output
FROM TBL
Output
223jjjj22
hYderabad
pUne
Here I fetched first character,upper case of second character and rest of characters separately. Then I concatenated them.
You can use similar approach for 3rd character if you want.
I have a cursor which selects date from column with NUMBER type containg floating point numbers. Numbers like 4,3433 are returned properly while numbers smaller then 1 have removed leading zero.
For example number 0,4513 is returned as ,4513.
When I execute select used in the cursor on the database, numbers are formatted properly, with leading zeros.
This is how I loop over the recors returned by the cursor:
FOR c_data IN cursor_name(p_date) LOOP
...
END LOOP;
Any ideas why it works that way?
Thank you in advance.
You're confusing number format and number value.
The two strings 0.123 and .123, when read as a number, are mathematically equals. They represent the same number. In Oracle the true number representation is never displayed directly, we always convert a number to a character to display it, either implicitly or explicitly with a function.
You assume that a number between 0 and 1 should be represented with a leading 0, but this is not true by default, it depends on how you ask this number to be displayed. If you don't want unexpected outcome, you have to be explicit when displaying numbers/dates, for example:
to_char(your_number, '9990.99');
It's the default number formatting that Oracle provides.
If you want to specify something custom, you shall use TO_CHAR function (either in SQL query or PL/SQL code inside the loop).
Here is how it works:
SQL>
SQL> WITH aa AS (
2 select 1.3232 NUM from dual UNION ALL
3 select 1.3232 NUM from dual UNION ALL
4 select 332.323 NUM from dual UNION ALL
5 select 0.3232 NUM from dual
6 )
7 select NUM, to_char(NUM, 'FM999990D9999999') FORMATTED from aa
8 /
NUM FORMATTED
---------- ---------------
1.3232 1.3232
1.3232 1.3232
332.323 332.323
.3232 0.3232
SQL>
In this example, 'FM' - suppresses extra blanks, '0' indicates number digit including leading/trailing zeros, and '9' indicates digit suppressing leading/trailing zeros.
You can find many examples here:
http://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements004.htm#i34570