I got this string /uk-en/contact-us/frequently-asked-questions/your-trip/there-wi-fi-access-in-the-eurostar-terminals-and-board-your-trains and I need to get just the last part of the URL (until the last /).
Then I want to replace '-' with a space. The strings are not with the same number of characters.
How can I do?
Thank you!
Solution using BigQuery functions:
select regexp_replace(last(split(x, "/")), "-", " ") from
(select
"/uk-en/contact-us/frequently-asked-questions/your-trip/there-wi-fi-access-in-the-eurostar-terminals-and-board-your-trains"
as x)
Here is what I tried in SQL Server
DECLARE #s VARCHAR(max)= '/uk-en/contact-us/frequently-asked-questions/your-trip/there-wi-fi-access-in-the-eurostar-terminals-and-board-your-trains'
SELECT REVERSE(SUBSTRING(REVERSE(#s),CHARINDEX('/',REVERSE(#s)),LEN(REVERSE(#s))))+REVERSE(REPLACE(SUBSTRING(REVERSE(#s),1,CHARINDEX('/',REVERSE(#s))-1),'-',' '))
Sorry this was for SQL Server
did you try using split in big query
SPLIT('str' [, 'delimiter']) Returns a set of substrings as a repeated string. If delimiter is specified, the SPLIT function breaks str into substrings, using delimiter as the delimiter.
Related
I have to trim Japanese characters string which has double byte space at start of string and end of string.
I have to do this by procedure of SQL server 2016.
For Example,
SELECT LTRIM(RTRIM(' A A '))
above one is working perfect
But Problem is in bellows line
SELECT LTRIM(RTRIM(' A A '))
i want output of above one is 'A A'
Have any idea, how to do this ?
Adapted SQL from OP's post:
SELECT LTRIM(RTRIM(REPLACE(' A A ', ' ', ' ')))
Screenshot with result:
The space in that string is the Ideographic space (U+3000) Unicode character, which LTRIM and RTRIM don't recognize as whitespace. Even TRIM in SQL Server 2017 won't recognize it unless it's specified explicitly.
Another problem is that this character is outside the normal range of characters and can't appear in a varchar field or value. This leads to inconsistent results between SQL Server versions. In SQL Server 2014 it will even appear as a ?. In later versions LTRIM/RTRIM may or may not work without emitting the error character. I don't have access to all versions to test this.
In SQL Server 2017 it's possible to explicitly specify the trimmed character, eg :
select trim(N' ' from N' A A ')
This produces A A.
In previous versions, PATINDEX can be used to find the locations of the first and last non-space positions :
declare #str nvarchar(10)=N' A A ';
declare #start int=PATINDEX(N'%[^ ]%',#str)
declare #end int=PATINDEX(N'% ',#str)
SELECT SUBSTRING(#str,#start,#end-#start)
The pattern N'%[^ ]%' finds the first non-U+3000 character in the string. N'% ' finds the position of the last one. SUBSTRING(#str,#start,#end-#start) extracts the content between the two positions.
The result is:
A A
I got solution
Thank you so much for your efforts.
Please use this function for double byte space remove.
CREATE FUNCTION [RTRIMBYTE](#AV_VALUE NVARCHAR(MAX))
RETURNS NVARCHAR(MAX)
AS
BEGIN
DECLARE #AV_RETURN NVARCHAR(MAX) = #AV_VALUE;
WHILE DATALENGTH(#AV_RETURN) > 0 AND RIGHT(#AV_RETURN, 1) in (' ', ' ')
SET #AV_RETURN = LEFT(#AV_RETURN, LEN('X' + #AV_RETURN + 'X') -3 ) ;
RETURN #AV_RETURN;
END;
I have a string as:
https://maps.googleapis.com/maps/api/staticmap?center=41.892532+-87.63811&zoom=11&scale=2&size=280x320&maptype=roadmap&format=png&visual_refresh=true%7C&markers=size:mid%7Ccolor:0x8000ff%7Clabel:1%7C2413+S+State+St++Chicago+IL+60616%7C&markers=size:mid%7Ccolor:0x8000ff%7Clabel:2%7C3000+N+Halsted+St++Chicago+IL+60657%7C&markers=size:mid%7Ccolor:0x8000ff%7Clabel:3%7C++++&key=AIzaSyBNEAQcC5niAEeiP3zkA_nuWGvtl0IOEs4
I want to replace the '++++' pattern at the end with blank and not the single occurrence of '+'. Tried using regexp_replace and translate functions in hive but that replaces all the single occurrences of '+' as well.
Use
regexp_replace(string,'[+]{4}','')
Pattern '[+]{4}' means + caracter four times.
Test:
select regexp_replace('++markers=size:mid%7Ccolor:0x8000ff%7Clabel:3%7C++++&','[+]{4}','');
Result:
OK
++markers=size:mid%7Ccolor:0x8000ff%7Clabel:3%7C&
Dod you try this?
replace(string, '++++', '')
Admittedly, this will replace all occurrences of '++++', but your string only has one of them.
I'm trying to extract everything after the first instance of a delimiter.
For example:
01443-30413 -> 30413
1221-935-5801 -> 935-5801
I have tried the following queries:
select regexp_replace(car_id, E'-.*', '') from schema.table_name;
select reverse(split_part(reverse(car_id), '-', 1)) from schema.table_name;
However both of them return:
01443-30413 -> 30413
1221-935-5801 -> 5801
So it's not working if delimiter appears multiple times.
I'm using Postgresql 11. I come from a MySQL background where you can do:
select SUBSTRING(car_id FROM (LOCATE('-',car_id)+1)) from table_name
Why not just do the PG equivalent of your MySQL approach and substring it?
SELECT SUBSTRING('abcdef-ghi' FROM POSITION('-' in 'abcdef-ghi') + 1)
If you don't like the "from" and "in" way of writing arguments, PG also has "normal" comma separated functions:
SELECT SUBSTR('abcdef-ghi', STRPOS('abcdef-ghi', '-') + 1)
I think that regexp_replace is appropriate, but using the correct pattern:
select regexp_replace('1221-935-5801', E'^[^-]+-', '');
935-5801
The regex pattern ^[^-]+- matches, from the start of the string, one or more non dash characters, ending with a dash. It then replaces with empty string, effectively removing this content.
Note that this approach also works if the input has no dashes at all, in which case it would just return the original input.
Use this regexp pattern :
select regexp_replace('1221-935-5801', E'^[^-]+-', '') from schema.table_name
Regexp explanation :
^ is the beginning of the string
[^-]+ means at least one character different than -
...until the - character is met
I tried it in a conventional way in general what we do (found
something similar to instr as strpos in postgrsql .) Can try the below
SELECT
SUBSTR(car_id,strpos(car_id,'-')+1,
length(car_id) ) from table ;
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)
Is it possible to delete part of string using regexp (or something else, may be something like CHARINDEX could help) in SQL query?
I use MS SQL Server (2008 most likely).
Example: I have strings like "[some useless info] Useful part of string" I want to delete parts with text in brackets if they are in line.
Use REPLACE
for example :
UPDATE authors SET city = replace(city, 'To Remove', 'With BLACK or Whatever')
WHERE city LIKE 'Salt%'; // with where condition
You can use the PATINDEX function. Its not a complete regular expression implementation but you can use it for simple things.
PATINDEX (Transact-SQL)> Returns the starting position of the first occurrence of a pattern in a specified expression, or zeros if the pattern is not found, on all valid text and character data types.
OR You can use CLR to extend the SQL Server with a complete regular expression implementation.
SQL Server 2005: CLR Integration
SELECT * FROM temp where replace(replace(replace(url,'http://',''),'www.',''),'https://','')='"+url+"';
You can use STUFF to insert a string into another string. It deletes a specified length of characters in the first string at the start position and then inserts the second string into the first string at the start position.
For example, the code below, replaces the 5 with 666666:
DECLARE #Variable NVARCHAR(MAX) = '12345678910'
SELECT STUFF(#Variable, 5, 1, '666666')
Note, that the second argument is not a string, it is a position and you are able to calculate it position using CHARINDEX for example.
Here is your case:
DECLARE #Variable NVARCHAR(MAX) = '[some useless info] Useful part of string'
SELECT STUFF(
#Variable
,CHARINDEX('[', #Variable)
,LEN(SUBSTRING(#Variable, CHARINDEX('[', #Variable), CHARINDEX(']', #Variable) - LEN(SUBSTRING(#Variable, 0, CHARINDEX('[', #Variable)))))
,''
)
Finally helps REPLACE, SUBSTRING and PATINDEX.
REPLACE(t.badString, Substring(t.badString , Patindex('%[%' , t.badString)+1 , Patindex('%]%' , t.badString)), '').
Thanks to all.