Keep substring that precedes an expression in SQLite - sql

I want to split a varchar column on a certain expression and keep the left hand side of the result.
My column looks as follows:
varchar_col
keep_this__discard_this
keep_this_too__discard_this
I want to split all the strings on the double underscore ('__') and keep whatever comes before it. How can this be done in SQLite?

You can use:
select substr(varchar_col, 1, instr(varchar_col, '__') - 1)
Here is a db<>fiddle.

Related

Substring of a specific occurence

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.

SQLite TRIM same character, multiple columns

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.

Cut string after first occurrence of a character

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)

Using SQL to make specific changes in a database.

I am trying to figure out some commands/code in SQL.
I have database with names, addresses IDs etc, but I have to convert firstname values ending in “jnr” to “(Jnr)” and those ending in “snr” to “(Snr)”.
How do I do this?
update table TABLE_NAME set NAMES = '*xyz*Jnr' where NAMES like '%jnr'
Update or select:
PASTE(column, CHAR_LENGTH(column)-3, 1, UPPER(SUBSTRING(column FROM CHAR_LENGTH(column)-3 FOR 1)
WHERE column LIKE '%jnr' OR column LIKE '%snr'
PASTE is used to put in one character at position 3 from end,
CHAR_LENGTH to get length of column value,
UPPER converts character to upper case,
SUBSTRING is used to pick one character here (j or s),
LIKE is used to find values ending with jnr, or snr.
All ANSI SQL (no dbms specified!)

SQL: insert space before numbers in string

I have a nvarchar field in my table, which contains all sorts of strings.
In case there are strings which contain a number following a non-number sign, I want to insert a space before that number.
That is - if a certain entry in that field is abc123, it should be turned into abc 123, or ab12.34 should become ab 12. 34.I want this to be done throughout the entire table.
What's the best way to achieve it?
You can try something like that:
select left(col,PATINDEX('%[0-9]%',col)-1 )+space(1)+
case
when PATINDEX('%[.]%',col)<>0
then substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[.]%',col))
+space(1)+
substring(col,PATINDEX('%[.]%',col)+1,len(col)+1-PATINDEX('%[.]%',col))
else substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[0-9]%',col))
end
from tab
It's not simply, but I hope it will help you.
SQL Fiddle
I used functions (link to MSDN):
LEFT, PATINDEX, SPACE, SUBSTRING, LEN
and regular expression.