I am looking for a way to extract all substrings between two delimiters over an entire column.I have found ways to do this for each string separately, but I need something I can apply over the entire column.
For example if I have a column called "NAMES" that contains the below values:
1235_brandon_098410090
1242353_sam_1920420101222
134214_kristein_39402384
I want my output to be
brandon
sam
kristein
how do I do this?
I've tried this:
regex_substr(names,'_(.*?)_'
Query Error: error: function regex_substr(character varying, unknown) does not exist
In MySQL, I think you can use substring_index():
select substring_index(substring_index(names, '_', 2), '_', -1)
This extracts the second value delimited by underscores, which is what all the sample data suggests is needed.
EDIT:
Your error message looks like Postgres. This is the equivalent in Postgres:
select v.*, split_part(names, '_', 2)
from (values ('1242353_sam_1920420101222')) v(names);
In Postgres, you can use substring() with a pattern as well:
select v.*, substring(names from '[A-Za-z]+')
Related
I have a column as varchar2 datatype, the data in it is in format:
100323.3819823.222
100.323123.443422
1001010100.233888
LOL12333.DDD33.44
I need to remove the whole part after the first occurrence of '.'
In the end it should look like this:
100323
100
1001010100
LOL12333
I cant seem to find the exact substring expression due to the fact that there is not any fix length of the first part.
One way is to use REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR(column_name,'^[^.]*') FROM table
The other way is to combine SUBSTR with INSTR, which is a bit faster, but will result in NULL if the data doesn't contain a dot, so you'll have to add a switch if needed:
SELECT SUBSTR(column_name, 1, INSTR(column_name,'.') - 1) FROM table
For oracle you can try this:
select substr (i,1,Instr(i,'.',i)-1) from Table name.
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 have a table in an SQLite db which has multiple columns with leading '='. I understand that I can use...
SELECT TRIM(`column1`, '=') FROM table;
to clean one column however I get a syntax error if I try for example, this...
SELECT TRIM(`column1`, `column2`, `column3`, '=') FROM table;
Due to incorrect number of arguments.
Is there a more efficient way of writing this code than applying the trim to each column separately like this?
SELECT TRIM(`column1`,'=')as `col1`, TRIM(`column2`,'=')as `col2`, TRIM(`column3`,'=')as `col3` FROM table;
How SQLite guide tells:
trim(X,Y)
The trim(X,Y) function returns a string formed by removing any and all
characters that appear in Y from both ends of X. If the Y argument is
omitted, trim(X) removes spaces from both ends of X.
You have only two parameters, so it's impossible apply it one shot on 3 columns table.
The first parameter is a column, or variable on you can apply trim. The second parameter is a character to change.
I have a column with assignment numbers like - 11827,27266,91717,09818-2,726252-3,8716151-0,827272,18181
Now i am selecting the records like
select assignment_number from table;
But now i want that the column detail is retreived in such a way that numbers are only retrieved without -2 -3 etc like
726252-3---> 726252 8716151-0-->8716151
I know i can use regex for this but i do not know how to use it
This will select everthing before the character -:
^([^-]+)
From 726252-3 will match 726252
You would use regexp() substr:
select regexp_substr(assignmentnumber, '[0-9]+')
This will return the first string of numbers encountered in the string.
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)