I'm trying to take the last four characters only from a varchar field. All the rows are different lengths. What function should I be using to accomplish this?
Right should do:
select RIGHT('abcdeffff',4)
SUBSTR(column, LENGTH(column) - 3, 4)
LENGTH returns length of string and SUBSTR returns 4 characters from "the position length - 4"
RIGHT ( character_expression , integer_expression )
SELECT RIGHT(column, 4) FROM ...
Also a list of other string functions.
Use the RIGHT() function: http://msdn.microsoft.com/en-us/library/ms177532(v=sql.105).aspx
SELECT RIGHT( '1234567890', 4 ); -- returns '7890'
For Oracle SQL, SUBSTR(column_name, -# of characters requested) will extract last three characters for a given query. e.g.
SELECT SUBSTR(description,-3) FROM student.course;
You can select last characters with -
WHERE SUBSTR('Hello world', -4)
I have Used RIGHT function in SQL Server and it's working.
SELECT RIGHT( your_column_name, 4 );
--It will show last 4 digit/char
You can just use
SELECT RIGHT(<Column_name>, 4) FROM <table_name>
This will be giving you all the lines but only with their last 4 characters.
tested solution on hackerrank....
select distinct(city) from station
where substr(lower(city), length(city), 1) in ('a', 'e', 'i', 'o', 'u') and substr(lower(city), 1, 1) in ('a', 'e', 'i', 'o', 'u');
Related
SQL....On my table I have attribute table with “Pat0700-1700” on my report I want to drop the Pat and only display 0700-1700. How would I accomplish this on SQL. I have search and tried the substring with neg results.
On these following RDBMS:
Oracle
MySQL
DB2
StandardSQL
you can try with the function SUBSTR():
SELECT SUBSTR(<column>, 4) AS substr_string
FROM <table>
OUTPUT:
substr_string
-------------
0700-1700
The standard SQL method would be replace():
select replace(col, 'Pat', '')
Given that the rest of the string has a fixed format -- 9 characters -- you might also find that one of these is appropriate (and more general):
select right(col, 9)
select substr(col, 4, 9) -- or perhaps substring()
I have value like the below format,
150000005705
160000004559
I would like to remove three zeros after first two digits
I tried with this SQL statement:
SELECT
'150000005705',
REPLACE('150000005705', SUBSTR('150000005705', 3, 3)) new_num
FROM DUAL
Output should be like
150005705
160004559
Use SUBSTR twice and concatenate the strings
SELECT '150000005705',
CONCAT(SUBSTR ('150000005705', 1, 2), SUBSTR ('150000005705', 6)) new_num
FROM DUAL
an alternative is to use replace if there are always 6 zero's
SELECT '150000005705',
REPLACE ('150000005705', '000000', '000') new_num
FROM DUAL
I'm trying to get the max value from a text field. All but two of the values are numbers with a single decimal. However, two of the values have something like 8.2.10. How can I pull back just the integer value? The values can go higher than 9.n, so I need to convert this field into a number so that I can get the largest value returned. So all I want to get back is the 8 from the 8.2.1.
Select cast(VERSION as int) is bombing out because of those two values with a second . in them.
You may derive by using regexp_substr with \d pattern :
with tab as
(
select regexp_substr('8.2.1', '\d', 1, 1) from dual
union all
select regexp_substr('9.0.1', '\d', 1, 1) from dual
)
select * from tab;
For Oracle you must attend the value as string for retire only the part before the dot. Ex:
SELECT NVL( SUBSTR('8.2.1',0, INSTR('8.2.1','.')-1),'8.2.1') AS SR FROM DUAL;
Check than the value is repeated 3 times in the sentence, and if the value is zero or the value didn't have decimal part then it will return the value as was set.
I had to use T-SQL rather PL/SQL, but the idea is the same:
DECLARE #s VARCHAR(10);
SELECT #s='8.2.1';
SELECT CAST(LEFT(#s, CHARINDEX('.', #s) - 1) AS INT);
returns the integer 8 - note that it won't work if there are no dots because it takes the part of the string to the left of the first dot.
If my quick look at equivalent functions was correct, then in Oracle that would end up as:
SELECT CAST(SUBSTR(VERSION, 1, INSTR(VERSION, '.') - 1) AS INT)
It may be very simple question, but I have run out of ideas.
I would like to remove first 5 characters from string.
Example string will look like:
1Y40K100R
I would like to display only digits that are after '%K' which in this case should give me result of 100R.
Please note that number after 'K' can have different amount of digits. It can be 4 digit number or 2 digit number.
Just use substr():
select substr(col, 6)
This returns all characters starting at the sixth.
There are multiple ways to return all characters after the k. If you know the string has a k, then use instr():
select substr(col, instr(col, 'K') + 1)
You can use regexp_substr
select regexp_substr('1Y40K100R', '(K)(.*)', 1, 1, 'i', 2) from dual
A way without regexp:
select substr('1Y40K100R', instr('1Y40K100R', 'K') +1) from dual
This may appear not so elegant, but it usually performs better than the regexp way.
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;