How to write sql query which will show values skipping first character if it is 0 (only the first character). All values are 3 characters long.
Examples:
numbers
123
023
003
102
should display as follows (after executing the query)
numbers
123
23
03
102
I used the following solution, but it removes all 0's, not just the first. How to fix it so that it only removes the first character if it is 0.
SUBSTRING(numbers, PATINDEX('%[^0]%', numbers+'.'), LEN(numbers))
I will be grateful for your help.
You can use CASE expression:
SELECT CASE WHEN LEFT(numbers, 1) = '0' THEN RIGHT(numbers, 2) ELSE numbers END AS FormattedNumbers
why not using simple substr() ?
select case when substr(mycol,1,1)='0' then substr(mycol,2) else mycol end
from my table
you did not mention your DB so i assumed its oracle. This will work in any RDBMS.
You can use charindex and substring methods to do string manipulation :)
select
case when charindex('0', number) = 1
then substring(number, 2, len(number))
else number end
from (
select '123' number
union all
select '023'
union all
select '003'
union all
select '102'
) a
Related
I am trying to remove the last n characters from a string. I tried this:
replace( str, right(str, 3), '' )
But it fails on str where the pattern repeats more than once. 888106106. In this case I get 888, instead of 888106
Now I am using
left (str, length(str)-3)
Is there a more efficient away of achieving this?
If you fancy a regex based solution:
regexp_replace(str,'...$','')
It will leave strings with < 3 characters unchanged
So checking that LEFT and/or SUBSTR work equally (I assume LEFT is faster):
select
column1
,left(column1, length(column1) -3) as r1
,substr(column1, 0, length(column1) -3) as r2
from values
('abc123')
,('ab123')
,('a123')
,('123')
,('12')
,('1')
,('')
,(null);
gives:
COLUMN1
R1
R2
abc123
abc
abc
ab123
ab
ab
a123
a
a
123
null
null
12
null
null
1
null
null
''
null
null
null
null
null
so no checks are needed, nice to know.
if you do some perf testing:
create database test;
create schema test.test;
create or replace table test.test.many_string as
select seq8()::text as a
from table(generator(ROWCOUNT => 10000000));
ALTER SESSION SET USE_CACHED_RESULT = false;
select sum(length(left(a, length(a) -3))) from test.test.many_string;
select sum(length(substr(a, 0, length(a) -3))) from test.test.many_string;
after running them both a couple of times on my x-small, I get results in the order of 300ms, so these are equal.
So it seems you have a fast solution, and easy to read.
In SQL Server, that's the most effective way.
Note that you should do one of:
assert that all input strings will be length>2 [a bit lazy]
handle the error where 1 of the rows has a length<3 and the query terminates early [a bit shoddy]
use a case statement to handle the case where length < 3 [the preferred approach]
CASE
WHEN LENGTH(str) > 2 THEN LEFT(str, LENGTH(str) - 3)
ELSE str
END
Other flavours of SQL you may have to work without the case statement.
I want to select columns in Oracle: Code and Number based on the condition,
If number is any of the following return “E”: ###7890001 through ###7890999
where #(first 3 digits) could be any digit. These numbers are saved as string values so I need to cast them in appropriate data type too.
The data returned would look like this:
Code Number
E 2347890001
E 9567890456
E 5647890999
Thanks in adavnce!!
Assuming your problem description is correct, your inputs are always 10-digit strings, and you must return code E if and only if the four digits in the middle are 7890. Right? If so, the simplest solution is to use the LIKE condition. Note that number is a reserved keyword, so it can't be used as column name (I hope your column name is not really number, is it?)
The string in the LIKE condition is hard to read; it consists of three underscore characters, then 7890, then three more underscore characters. The underscore stands for exactly one character (any character).
with
simulated_data (num) as (
select '1234567890' from dual union all
select '3337890456' from dual union all
select '7897890455' from dual union all
select '9998887774' from dual
)
select num, case when num like '___7890___' then 'E' end as code
from simulated_data
order by num
;
NUM CODE
---------- ----
1234567890
3337890456 E
7897890455 E
9998887774
You can use to_number with mod and on conversion error (oracle 12.2 or higher) as following:
Select 'E' as code,
number
From your_table
Where MOD(TO_NUMBER(number DEFAULT -1 ON CONVERSION ERROR), 10000000)
between 7890001 and 7890999;
Cheers!!
This should work:
where code = 'E' and
substr(number, -7) between '7890001' and '7890999'
EDIT:
select (case when substr(number, -7) between '7890001' and '7890999' then 'E' end) as code
I need to replace the entire word with 0 if the word has any non-digit character. For example, if digital_word='22B4' then replace with 0, else if digital_word='224' then do not replace.
SELECT replace_funtion(digital_word,'has non numeric character pattern',0,digital_word)
FROM dual;
I tried decode, regexp_instr, regexp_replace but could not come up with the right solution.
Please advise.
Thank you.
the idea is simple - you need check if the value is numeric or not
script:
with nums as
(
select '123' as num from dual union all
select '456' as num from dual union all
select '7A9' as num from dual union all
select '098' as num from dual
)
select n.*
,nvl2(LENGTH(TRIM(TRANSLATE(num, ' +-.0123456789', ' '))),'0',num)
from nums n
result
1 123 123
2 456 456
3 7A9 0
4 098 098
see more articles below to see which way is better to you
How can I determine if a string is numeric in SQL?
https://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:15321803936685
How to tell if a value is not numeric in Oracle?
You might try the following:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '\D') THEN '0' ELSE digital_word END
FROM dual;
The regular expression class \D matches any non-digit character. You could also use [^0-9] to the same effect:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '\D') THEN '0' ELSE digital_word END
FROM dual;
Alternately you could see if the value of digital_word is made up of nothing but digits:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '^\d+$') THEN digital_word ELSE '0' END
FROM dual;
Hope this helps.
The fastest way is to replace all digits with null (to simply delete them) and see if anything is left. You don't need regular expressions (slow!) for this, you just need the standard string function TRANSLATE().
Unfortunately, Oracle has to work around their own inconsistent treatment of NULL - sometimes as empty string, sometimes not. In the case of the TRANSLATE() function, you can't simply translate every digit to nothing; you must also translate a non-digit character to itself, so that the third argument is not an empty string (which is treated as a real NULL, as in relational theory). See the Oracle documentation for the TRANSLATE() function. https://docs.oracle.com/cd/E11882_01/server.112/e41084/functions216.htm#SQLRF06145
Then, the result can be obtained with a CASE expression (or various forms of NULL handling functions; I prefer CASE, which is SQL Standard):
with
nums ( num ) as (
select '123' from dual union all
select '-56' from dual union all
select '7A9' from dual union all
select '0.9' from dual
)
-- End of simulated inputs (for testing only, not part of the solution).
-- SQL query begins BELOW THIS LINE. Use your own table and column names.
select num,
case when translate(num, 'z0123456789', 'z') is null
then num
else '0'
end as result
from nums
;
NUM RESULT
--- ------
123 123
-56 0
7A9 0
0.9 0
Note: everything here is in varchar2 data type (or some other kind of string data type). If the results should be converted to number, wrap the entire case expression within TO_NUMBER(). Note also that the strings '-56' and '0.9' are not all-digits (they contain non-digits), so the result is '0' for both. If this is not what you needed, you must correct the problem statement in the original post.
Something like the following update query will help you:
update [table] set [col] = '0'
where REGEXP_LIKE([col], '.*\D.*', 'i')
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, '.*(\..*)$', 1, 1, NULL, 1), '[^0-9\.]+', '') Converted
from tbl;
SQL> /
ORIGINAL CONVERTED
--------------- ---------------
3,000,000.000 3000000.000
3.000.000.000 3000000.000
a^*3^%*(C4.5d*9 34.59
about 1,000.00 1000.00
SQL>
Shortest way is as follows:
select regexp_substr('a^*3^%*(C4.5d*9s','\d+\.\d+') from dual;
or
select regexp_replace('a^*3^%*(C4.5d*9s', '[^0.0-9]', '') from dual;
I have to select only the IDs which have only even digits (an ID looks like: p19 ,p20 etc). That is, p20 is good (both 2 and 0 are even digits); p18 is not.
I thought to use substr to get each number from the IDs and then see if it's even .
select from profs
where to_number(substr(id_prof,2,2))%2=0 and to_number(substr(id_prof,3,2))%2=0;
IF you need all rows consist of 'p' in beginning and even digits on tail It should look like:
select *
from profs
where regexp_like (id_prof, '^p[24680]+$');
with
profs ( prof_id ) as (
select 'p18' from dual union all
select 'p24' from dual union all
select 'p53' from dual
)
-- End of test data; what is above this line is NOT part of the solution.
-- The solution (SQL query) begins here.
select *
from profs
where length(prof_id) = length(translate(prof_id, '013579', '0'));
PROF_ID
-------
p24
This solution should work faster than anything using regular expressions. All it does is to replace 0 with itself and DELETE all odd digits from the input string. (The '0' is included due to a strange but documented behavior of translate() - the third argument can't be empty). If the length of the input string doesn't change after the translation, that means the input string didn't have any odd digits.
where mod(to_number(regexp_replace(id_prof, '[^[:digit:]]', '')),2) = 0