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)
Related
I am trying to split a field by delimiter in LookML. This field either follows the format of:
Managers (AE)
Managers (AE - MM)
I was able to split to first case using this
sql: case
when rlike (${user_role_name}, '^.*[\\(\\)].*$') then split_part(${user_role_name}, ' ', -1)
However, I haven't been able to get the 2nd case to do the same. It's in a case statement so I am going to add another when statement, but am not able to figure out the regex for parentheses that contains spaces.
Thanks in advance for the help!
By "split" the string, I think you mean you want to extract the part in parentheses, right?
I would do this using a regex substring method. You didn't mention what warehouse you're using, and the syntax will vary a little, but on snowflake that would look like:
regexp_substr(${user_role_name}, '\\([^)]*\\)')
So, for example, with the inputs you gave:
select regexp_substr('Managers (AE)', '\\([^)]*\\)')
union all
select regexp_substr('Managers (AE - MM)', '\\([^)]*\\)')
result
(AE)
(AE - MM)
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 ;
How to ignore special characters and get only number with the below input as string.
Input: '33-01-616-000'
Output should be 3301616000
Use the REPLACE() function to remove the - characters.
REPLACE(columnname, '-', '')
Or if there can be other non-numeric characters, you can use REGEXP_REPLACE() to remove anything that isn't a number.
REGEXP_REPLACE(columnname, '\D', '')
Standard string functions (like REPLACE, TRANSLATE etc.) are often much faster (one order of magnitude faster) than their regular expression counterparts. Of course, this is only important if you have a lot of data to process, and/or if you don't have that much data but you must process it very frequently.
Here is one way to use TRANSLATE for this problem even if you don't know ahead of time what other characters there may be in the string - besides digits:
TRANSLATE(columnname, '0123456789' || columnname, '0123456789')
This will map 0 to 0, 1 to 1, etc. - and all other characters in the input string columnname to nothing (so they will be simply removed). Note that in the TRANSLATE mapping, only the first occurrence of a character in the second argument matters - any additional mapping (due to the appearance of the same character in the second argument more than once) is ignored.
You can also use REGEXP_REPLACE function. Try code below,
SELECT REGEXP_REPLACE('33-01-61ASDF6-0**(98)00[],./123', '([^[:digit:]])', NULL)
FROM DUAL;
SELECT regexp_replace('33-01-616-000','[^0-9]') digits_only FROM dual;
/
Input string: ["1189-13627273","89-13706681","118-13708388"]
Expected Output: ["14013627273","14013706681","14013708388"]
What I am trying to achieve is to replace any numbers till the '-' for each item with hard coded text like '140'
SELECT replace(value_to_replace, '-', '140')
FROM (
VALUES ('1189-13627273-77'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
check this
I found the right way to achieve that using the below regular expression.
SELECT REGEXP_REPLACE (string_to_change, '\\"[0-9]+\\-', '140')
You don't need a regexp for this, it's as easy as concatenation of 140 and the substring from - (or the second part when you split by -)
select '140'||substring('89-13706681' from position('-' in '89-13706681')+1 for 1000)
select '140'||split_part('89-13706681','-',2)
also, it's important to consider if you might have instances that don't contain - and what would be the output in this case
Use regexp_replace(text,text,text) function to do so giving the pattern to match and replacement string.
First argument is the value to be replaced, second is the POSIX regular expression and third is a replacement text.
Example
SELECT regexp_replace('1189-13627273', '.*-', '140');
Output: 14013627273
Sample data set query
SELECT regexp_replace(value_to_replace, '.*-', '140')
FROM (
VALUES ('1189-13627273'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
Caution! Pattern .*- will replace every character until it finds last occurence of - with text 140.
I am using this query to replace one character in a cell
select replace(id,',','')id from table
But I want to replace two characters in a cell.
If the cell is having this data (1,3.1), and I want it to look like this (131).
How can I replace two different characters in one cell?
Use TRANSLATE instead of REPLACE(). It replaces each occurrence of a character in the first pattern with its matched character in the second. To remove characters, simply leave cut short the replacement string:
select translate(id, '1,.', '1') id from table
Note that the second string cannot be null. Hence the need to include 1 (or some other character) in both strings.
Find out more.
Obviously the more characters you need to convert/remove the more attractive TRANSLATE() becomes. The main use for REPLACE is changing patterns (such as words) rather than individual characters.
Can use
select replace(translate(id,',.',' '),' ','') from table;
or
select regexp_replace('1,3.1','[,.]','') from dual;
or
select replace(replace(id,',',''),'.','') from table;
Call the replace again.
select replace(replace(id,',',''), '.','') id from table
Do this:
select REPLACE(REPLACE(id,',',''),'.','')
Or use a regular expression:
select regexp_replace(id, '[.,]', '') id from table
Find out more