Suppose there is a value 842545/003. I need to take the part after '/'.
84454/02. I want a query to take the only 02 from here
You can try below substr() and instr() function
select substr('84454/02',instr('84454/02','/')+1,
length('84454/02')-instr('84454/02','/')) as val
from dual
You can use a regex (with the usual warnings about regex performance - the simple string functions like instr and substr are faster if you are processing millions of rows).
regexp_replace(yourcolumn, '^.*/')
This removes everything up to and including the / character (or the final one if there is more than one).
If your are using SQL, you can use this.
SELECT SUBSTRING('84454/02',PATINDEX('%/%', '84454/02')+1,LEN('84454/02'));
'84454/02'= Column name
Related
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)
I have strings like 'keepme:cutme' or 'string-without-separator' which should become respectively 'keepme' and 'string-without-separator'. Can this be done in PostgreSQL? I tried:
select substring('first:last' from '.+:')
But this leaves the : in and won't work if there is no : in the string.
Use split_part():
SELECT split_part('first:last', ':', 1) AS first_part
Returns the whole string if the delimiter is not there. And it's simple to get the 2nd or 3rd part etc.
Substantially faster than functions using regular expression matching. And since we have a fixed delimiter we don't need the magic of regular expressions.
Related:
Split comma separated column data into additional columns
regexp_replace() may be overload for what you need, but it also gives the additional benefit of regex. For instance, if strings use multiple delimiters.
Example use:
select regexp_replace( 'first:last', E':.*', '');
SQL Select to pick everything after the last occurrence of a character
select right('first:last', charindex(':', reverse('first:last')) - 1)
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 require a select query that adds a space to the data based on the placement of the capital letters i.e. 'HelpMe' using this query would be displayed as 'Help Me' . Note i cannot use a stored function to do this the it must be done in the query itself. The Data is of variable length and query must be in SQL. Any Help will be appreciated.
Thanks
You need to use user defined function for this until MS give us support for regular expressions. Solution would be something like:
SELECT col1, dbo.RegExReplace(col1, '([A-Z])',' \1') FROM Table
Aldo this would produce leading space that you can remove with TRIM.
Replace regular expresion function:
http://connect.microsoft.com/SQLServer/feedback/details/378520
About dbo.RegexReplace you can read at:
TSQL Replace all non a-z/A-Z characters with an empty string
Assume if you are using Oracle RDBMS, you use the following,
REGEX_REPLACE
SELECT REGEXP_REPLACE('ILikeToWatchCSIMiami',
'([A-Z.])', ' \1')
AS RX_REPLACE
FROM dual
;
Managed to get this output: * SQLFIDDLE
But as you see it doesn't treat well on words such as CSI though.
I need to extract string from a value of a cell in Oracle table. How can I do that?
VALUE
--------
/var1/var2/var3/C751994ZP1QT11/var4.itp
I want to extract "C751994ZP1QT11" from the value using SQL. I searched some function like REGEXP_LIKE, but I am not quite understand how to use it.
Thanks ahead for the help.
If you want to extract a string, you need to use regexp_substr(). regexp_like() would return a boolean indicating if a certain regular expression matches your string, while regexp_substr() actually returns the matched string.
In your case, assuming you want the string between /** and **/, you could use something like this:
select
regexp_substr(
'/var1/var2/var3/**C751994ZP1QT11**/var4.itp',
'/\*\*.*\*\*/')
from dual
Refer here for some useful shortcuts on regular expressions with Oracle.
select REGEXP_REPLACE('/var1/var2/var3/**C751994ZP1QT11**/var4.itp','.*\*\*(.*)\*\*.*','\1') result from dual