I am removing that last 3 characters from the string "ABC123" using regexp_replace function in Oracle using the below statement
select REGEXP_REPLACE('ABC123','123','', LENGTH('ABC123') - 3) from dual;
The same result can be achieved in Postgres with the below statements,
select regexp_replace('ABC123','[123]', '','g')
select translate('ABC123','123', '');
Is there any way I can use the length function for replace as I have used in Oracle?
Why not simply use left()?
select left('ABC123', length('ABC123') - 3)
The same idea can be used in Oracle as well, but you need to use the substr() function. This should be more efficient in both databases.
You could also look into the trim functionality.
http://www.postgresqltutorial.com/postgresql-trim-function/
"select REGEXP_REPLACE('ABC123','123','', LENGTH('ABC123') - 3) from dual;"
would become
select ltrim('ABC123','ABC') from dual;
resulting in 123
Related
Here is what i do in prestosql
select split_part('one|||two','|||',1)
result = 'one'
but I can't use this function(split_part) in HIVEsql
Is there any possible function for me to get the same result as above but in HIVEsql?
Maybe this way?
SELECT split('one|||two','[\\|||]')[0]
The "|" is a reserved character, so you need to scape it properly
Hive does have a split() function, but it returns an array. That would be:
select split('one|||two', '[|]{3}')[0]
What is the Syntax to substr in Oracle to subtract a string
i have "123456789 #073"
I only want what after the #
substr (table.col, 17,3)
is that ok ?
Most likely the simplest (and most performant) way of doing this would be to use the base string functions:
SELECT SUBSTR(col, INSTR(col, '#') + 1)
FROM yourTable;
Demo
We could also try using REGEXP_REPLACE here:
SELECT REGEXP_REPLACE(col, '.*#(.*)', '\1')
FROM yourTable;
The regex option would in general not perform as well as the first query. The reason for this is that invoking a regex incurs a performance overhead. You might want to consider a regex option if you expect that the string logic might change or get more complicated in the future. Otherwise, go with base string functions wherever possible.
I think the most direct method might be regexp_substr():
select regexp_substr('123456789 #073', '[^#]+$')
from dual;
The regular expression says: "get me all non-hash characters at the end of the string".
If you happen to know that there are 3 characters and really want the last three characters of the string:
select substr('123456789 #073', -3)
Table A
ID ID_Descr
1 'DUP 8002061286'
2 'DUP 8002082667 '
3 ' 8002082669 DUP'
I would like to extract the string from the ID_Descr field with the following conditions:
String always starts with 8
String length is always 10 digits
This means stripping everything to the right and left of the string (eg. '8002082669'). How can I achieve this? Using REGEXP_SUBSTR?
I am using Oracle 11g.
Thanks!
Although you could use regexp_substr() for this, I would take a different approach. I would just look for the '8' using instr() and then take the next 10 characters:
select substr(id_descr, instr(id_descr, '8'), 10)
This seems like the simplest solution.
You could use REGEXP_SUBSTR() but a regex is an expensive operation so you would be much better off using SUBSTR() and INSTR():
SELECT SUBSTR(ID_Descr, INSTR(ID_Descr, '8'), 10) FROM tableA;
If you really wanted to use REGEXP_SUBSTR() you could do it as follows:
SELECT REGEXP_SUBSTR(ID_Descr, '8.{9}') FROM tableA;
This would get 8 plus the following 9 characters (. being the wildcard).
Now if you wanted to match only digits, then REGEXP_SUBSTR() would probably be your best bet:
SELECT REGEXP_SUBSTR(ID_Descr, '8[0-9]{9}') FROM tableA;
I am using Oracle 11 G and have the following set of data:
12.0
4.2
Version.1
7.9
abc.72
I want to return all string characters before the period. What sort of query would I run in order to achieve this? Any help would be greatly appreciated, thanks!
You can try a combination of instr and substr.
Something like this:
select substr(field, 1, instr(field, '.') - 1)
from your_table;
Assuming field always contains a . character on it.
You can also deal with strings without a . by using case, if or any other similar valid conditional function on Oracle's SQL language implementation.
Of course, you can always put this on a function to make it look nicer on your query.
Lets say I have following string: 'product=1627;color=45;size=7' in some field of the table.
I want to query for the color and get 45.
With this query:
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?') "colorID"
FROM DUAL;
I get :
colorID
---------
color=45;
1 row selected
.
Is it possible to get part of the matched string - 45 for this example?
One way to do it is with REGEXP_REPLACE. You need to define the whole string as a regex pattern and then use just the element you want as the replace string. In this example the ColorID is the third pattern in the entire string
SELECT REGEXP_REPLACE('product=1627;color=45;size=7'
, '(.*)(color\=)([^;]+);?(.*)'
, '\3') "colorID"
FROM DUAL;
It is possible there may be less clunky regex solutions, but this one definitely works. Here's a SQL Fiddle.
Try something like this:
SELECT REGEXP_SUBSTR(REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?'), '[[:digit:]]+') "colorID"
FROM DUAL;
From Oracle 11g onwards we can specify capture groups in REGEXP_SUBSTR.
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color=(\d+);', 1, 1, 'i', 1) "colorID"
FROM DUAL;