How to use regex replace in Postgres function? - sql

I have postgres function in which i am appending values in query such that i have,
DECLARE
clause text = '';
after appending i have some thing like,
clause = "and name='john' and age='24' and location ='New York';"
I append above in where clause of the query i already have. While executing query i am getting "and" just after "where" result in error
How to use regex_replace so that i remove the first "and" from clause before appending it to the query ?

Instead of fixing clause after the fact, you could avoid the problem by using
concat_ws (concatenate with separator):
clause = concat_ws(' and ', "name='john'", "age='24'", "location ='New York'")
will make clause equal to
"name='john' and age='24' and location ='New York'"

This can be even simpler. Use right() with a negative offset.
Truncates the first n characters and you don't need to specify the length of the string. Faster, simpler.
Double quotes (") are for identifiers in Postgres (and standard SQL) and incorrect in your example. Enclose string literals in single quotes (') and escape single quotes within - or use dollar quoting:
Insert text with single quotes in PostgreSQL
Since this is a plpgsql assignment, use the proper assignment operator :=. The SQL assignment operator = is tolerated, too, but can lead to ambiguity in corner cases.
Finally, you can assign a variable in plpgsql at declaration time. Assignments in plpgsql are still cheap but more expensive than in other programming languages.
DECLARE
clause text := right($$and name='john' and age='24' ... $$, -5)
All that said, it seems like you are trying to work with dynamic SQL and starting off on the wrong foot here. If those values can change, rather supply them as values with the USING clause of EXECUTE and be wary of SQL injection. Read some of the related questions and answers on the matter:
https://stackoverflow.com/search?q=[plpgsql]+[dynamic-sql]+EXECUTE+USING

You do not need regex:
clause = substr(clause, 5, 10000);
clause = substr(clause, 5, length(clause)- 4); -- version for formalists

concat_ws sounds like the best option, but as a general solution for things like this (or any sort of list with a delimiter) you can use logic like (pseudocode):
delim = '';
while (more appendages)
clause = delim + nextAppendage;
delim = ' AND ';

If you want to do it with regular expression try this:
result = regexp_replace(clause, '^and ', '')

Related

Match Character Whether or Not It Exists in Like Statement

I need a like expression that will match a character whether or not it exists. It needs to match the following values:
..."value": "123456"...
..."value": "123456"...
"...value":"123456"...
This like statement will almost work: LIKE '%value":%"123456"%'
But there are values like this one that would also match, but I don't want returned:
..."value":"99999", "other":"123456"...
A regex expression to do what I'm looking to do is 'value": *?"123456"'. I need to do this in SQL Server 2008 and I don't believe there is good regex support in that version. How can I match using a like statement?
Remove the whitespace in your compare with REPLACE():
WHERE REPLACE(column,' ','') LIKE '%"value":"123456"%'
May need a double replace for tabs:
REPLACE(REPLACE(column,' ',''),' ','')
I don't think you can with the like operator. You could exclude ones you could match, like if you want to make sure it just doesn't contain other:
[field] LIKE '%value":%"123456"%` AND [field] NOT LIKE '%"other"%'
Otherwise I think you'd have to do some processing on the string. You could write a UDF to take the string and parse it to find the value for 'value' and compare based on that:
dbo.fn_GetValue([field], 'value') = '123456'
The function could find the index of '"' + #name + '"', find the next index of a quote, and the one after that, then get the string between those two quotes and return it.

Remove " " from values

In PostgreSQL a column contains values such as:
"Sample1"
"Sample2"
Is there an efficient way to go through each record removing the " " marks?
This seems like a relatively straightforward operation, but I am baffled.
Try this:
UPDATE TableName
SET ColName = REPLACE(ColName, '"', '');
WHERE ColName SIMILAR TO '%"%'
You can use LIKE operator as well instead of SIMILAR TO.
Since you asked for an efficient way:
UPDATE tbl
SET col = translate(col, '"', '')
WHERE col LIKE '%"%';
Never use SIMILAR TO, its an utterly pointless operator, only included in Postgres for standard compliance. There is hope the standard will include a regular expression match (like ~ in Postgres) instead in the future.
For replacing single characters, translate() is simpler and faster than replace().
If you only want to replace leading and / or trailing characters, use trim() / ltrim() / rtrim() instead. And also change to WHERE col LIKE '"%' OR col LIKE '%"' respectively.

How to remove part of the string in oracle

Input data:
abcdef_fhj_viji.dvc
Expected output:
fhj_viji.dvc
The part to be trimmed is not constant.
Use the REPLACE method
Select REPLACE('abcdef_fhj_viji.dvc','abcde','')
If you want this query for your table :
Select REPLACE(column,'abcde','') from myTable
For update :
UPDATE TABLE
SET column = REPLACE(column,'abcde','')
select substr('abcdef_fhj_viji.dvc',instr('abcdef_fhj_viji.dvc','_')+1) from dual
So, Its all depends on INSTR function, define from which position and which occurrence, you will get the index and pass that index to SUBSTR to get your string.
Since you didn't give a lot of information I'm gonna assume some.
Let's assume you want a prefix of some string to be deleted. A good way to do that is by using Regular Expressions. There's a function called regexp_replace, that can find a substring of a string, depending on a pattern, and replace it with a different string. In PL/SQL you could write yourself a function using regexp_replace, like this:
function deletePrefix(stringName in varchar2) return varchar2 is
begin
return regexp_replace(stringName, '^[a-zA-Z]+_', '');
end;
or just use this in plain sql like:
regexp_replace(stringName, '^[a-zA-Z]+_', '');
stringName being the string you want to process, and the ^[a-zA-Z]+_ part depending on what characters the prefix includes. Here I only included upper- and lowercase letters.

how to use a string that contain ' in SQL "IN" clause

i have a string value like 'Apple's'. i want to use this string in SQL "IN" clause like below query
select * from tbl_fruit where nm_fruit IN(''Apple's'','Orange');
how i can get the above query work correctly ?
Many Thanks,
Awais Afzal.
double the single quotes,
select * from tbl_fruit where nm_fruit IN ('Apple''s', 'Orange')
but if you do it on the application level, make sure you parameterized the query :)
I have found SQL correctly interprets the ASCII single-closed-quote (ALT 0146) as an apostrophe in searches while the "IN" treats it like any other character. Now I can search for 'Matt's Macintosh' using Matt(ASCII character 0146)s Macintosh" without messing up my list or the search.

Select query that displays Joined words separately, not using a function

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.