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.
Related
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
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')
Although I saw update statements to update field based on existing values, I could not find anything similar to this scenario:
Suppose you have a table with only one column of number(4) type. The value in the first record is 1010.
create table stab(
nmbr number(4)
);
insert into stab values(1010);
For each digit
When the digit is 1 -- add 3 to the digit
When the digit is 0 -- add four to the digit
end
This operations needs to be completed in a single statement without using pl/sql.
I think substr function need to be used but don't know how to go about completing this.
Thanks in advance.
SELECT DECODE(SUBSTR(nmbr,1,1), '1', 1 + 3, '0', 0 + 4) AS Decoded_Nmbr
FROM stab
ORDER BY Decoded_Nmbr
Is that what you are after?
So, it seems you need to convert every 0 and 1 to a 4, and leave all the other digits alone. This seems like a string operation (and the reference to "digits" itself suggests the same thing). So, convert the number to a string, use the Oracle TRANSLATE function (see the documentation), and convert back to number.
update stab
set nmbr = to_number(translate(to_char(nmbr, '9999'), '01', '44'))
;
assuming its always a 4 digit #; you could use substring like below
-- postgres SQL example
SELECT CASE
WHEN a = 0 THEN a + 4
ELSE a + 3
end AS a,
CASE
WHEN b = 0 THEN b + 4
ELSE b + 3
end AS b,
CASE
WHEN c = 0 THEN c + 4
ELSE c + 3
end AS c,
CASE
WHEN d = 0 THEN d + 4
ELSE c + 3
end AS d
FROM ( SELECT Substr( '1010', 1, 1 ) :: INT AS a,
Substr( '1010', 2, 1 ) :: INT b,
Substr( '1010', 3, 1 ) :: INT c,
Substr( '1010', 4, 1 ) :: INT d )a
--- Other option may be (tried in postgreSQL :) ) to split the number using regexp_split_to_table into rows;then add individual each digit based on the case statement and then concat the digits back into a string
SELECT array_to_string ( array
(
select
case
WHEN val = 0 THEN val +4
ELSE val +3
END
FROM (
SELECT regexp_split_to_table ( '101010','' ) ::INT val
) a
) ,'' )
My answer to the interview question would have been that the DB design violates the rules of normalization (i.e. a bad design) and would not have this kind of "update anomaly" if it were properly designed. Having said that, it can easily be done with an expression using various combinations of single row functions combined with the required arithmetic operations.
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
oI have a teradata table with about 10 million records in it, that stores a numeric id field as a varchar. i need to transfer the values in this field to a bigint column in another table, but i can't simply say cast(id_field as bigint) because i get an invalid character error. looking through the values, i find that there could be a character at any position in the string, so let's say the string is varchar(18) i could filter out invalid rows like so :
where substr(id_field,1,1) not in (/*big,ugly array of non-numeric chars*/)
and substr(id_field,2,1) not in (/*big,ugly array of non-numeric chars*/)
etc, etc...
then the cast would work, but this is not feasible in the long run. it's slow and if the string has 18 possible characters, it makes the query unreadable. how can i filter out rows that have a value in this field that will not cast as a bigint without checking each character individually for an array of non-numeric characters?
example values would be
123abc464
a2.3v65
a_356087
........
000000000
BOB KNIGHT
1235468099
the values follow no specific patterns, I simply need to filter out the ones that contain ANY non-numeric data.
123456789 is okay but 123.abc_c3865 is not...
Starting with TD14 Teradata added some functions, now there are multiple ways, e.g.:
WHERE RTRIM(col, '0123456789') = ''
But the easiest way is TO_NUMBER, which returns NULL for bad data:
TO_NUMBER(col)
The best that I've ever managed is this:
where char2hexint(upper(id_field)) = char2hexint(lower(id_field))
Since upper case characters give a different hex value to lower case ones, this will ensure that you have no alphabetical characters, but will still leave you with underscores, colons and so forth. If this doesn't meet your requirements, you may need to write an UDF.
could we also try to divide the values in the field by some integer "if divided then must be a number and if not and throws some error,then must have some character...." guess this would be lot fast as has just mathematics involved...
I've faced the same issue to try to exclude alpha characters from street address house numbers. The following will work if you don't mind concatanating all the numeric numbers together......
It checks if the upper of a string equals the lower of the string, if so it's a number, if not it becomes null.
select cast(case when upper(substring('12E'from 1 for 1)) = lower(substring('12E'from 1 for 1)) then substring('12E'from 1 for 1) else null end ||
case when upper(substring('12E'from 2 for 1)) = lower(substring('12E'from 2 for 1)) then substring('12E'from 2 for 1) else null end ||
case when upper(substring('12E'from 3 for 1)) = lower(substring('12E'from 3 for 1)) then substring('12E'from 3 for 1) else null end ||
case when upper(substring('12E'from 4 for 1)) = lower(substring('12E'from 4 for 1)) then substring('12E'from 4 for 1) else null end ||
case when upper(substring('12E'from 5 for 1)) = lower(substring('12E'from 5 for 1)) then substring('12E'from 5 for 1) else null end ||
case when upper(substring('12E'from 2 for 1)) = lower(substring('12E'from 2 for 1)) then substring('12E'from 2 for 1) else null end
as integer)
Try using this code segment
WHERE id_Field NOT LIKE '%[^0-9]%'
I found lins314159 answer to be very helpful with a similar issue. It may be an old thread but for what it's worth, I used:
char2hexint(upper(id_field)) = char2hexint(lower(id_field)) AND substr(id_field,1,1) IN ('1' to '9')
to successfully cast the remaining VARCHAR results to INT
SELECT customer_id
FROM t
WHERE UPPER(customer_id)(CASESPECIFIC) <>
LOWER(customer_id)(CASESPECIFIC);
This works perfectly fine to check whether the values in a numeric field is non-numeric.
SELECT id_field
WHERE oTranslate(id_field, '0123456789','')<>'';
This works well for me! It reveals any id_field containing a non-numeric value