I am new to presto, I am looking to use regex_replace on a particular db column instead of a string.
E.g: Replace all entries from a column "Description" that starts with digit and followed by space from "table1"
Can someone please help with an example?
I tried this :
select(regexp_replace(col("Description"), '\d+\s')) from table1
but getting error: "Function not found", function "Col" is not registered.
Thanks in advance!
Its enough to put column name and a replacement string should be added. So,
col("Description") -> Description
-> ,'replace_string'
Final query should be something like this :
select regexp_replace(Description, '\d+\s','replace_string' ) from table1 b
I think this does what you want:
SELECT
REGEXP_REPLACE(`Description`, '\d+\s')
FROM `table1`
They're optional in SQL, but you use backticks for column and table names, apostrophes for strings.
Related
I want to change some data in the database where the text of a column is in a string.
For example we have the String: 'QWZ-$ §_REMARKIUFZQ'
Now I want to select a column where the column name has the text 'Mark'.
Something like this stupid try:
SELECT info FROM database WHERE '%'+name+'%' LIKE 'QWZ-$ §_REMARKIUFZQ'
Best Regards,
Christian
From sqlite doc:
The operand to the right of the LIKE operator contains the pattern and the left hand operand contains the string to match against the pattern.
Switch the LIKE operands i.e. WHERE 'QWZ-$ §_REMARKIUFZQ' LIKE '%'||name||'%' (notice + changed to sqlite concatenate operator || for native sqlite syntax).
Assuming sqlite3 version >= 3.16, you may find the table valued function pragma_table_info is the place to look.
I'm writing a query that returns a bunch of things from multiple tables. The main query is against Table_1. I need to return a substring from a field in table 7. But I'm getting an error that Substring_Index is an invalid identifier. How can I achieve the intended result?
I have a field COLUMN_1 of TABLE_1 that has 3+ pieces of data, separated by " : " (space colon space) and I need to strip out the text before the first delimiter, and return the rest of it (regardless of length).
A simplified example:
SELECT t1.name
,t1.address
,t1.phone
,t2. fave_brand
,SUBSTRING_INDEX(t3.fave_product, ' : ', -1) AS Fave Product
FROM table_1 t1
INNER JOIN table_2 t2
ON t2.brand_SK = t1.fave_brand_FK
INNER JOIN table_3 t3
ON t3.product_list_SK = t1.fave_products
WHERE <a series of constraints>;
Please note, I am NOT normally an SQL developer, but the back-end dev is on vacation and I've been tasked with cobbling this fix together. I'm a beginner at best.
In oracle you could use regexp_replace():
regexp_replace(t3.fave_product, '^[^:]*:', '') "Fave Product"
regexp_replace() replaces the part of the string that matches the regexp given as second argument with the value given as third argument. Here, we use the empty string as third argument, meaning that the matching part of the string is suppressed.
Regexp breakdown:
^ beginning of the string
[^:]* as many characters as possible other than ":" (possibly, 0 characters)
: character ":"
NB: identifiers that contain special characters (such as space) need to be double quoted.
Oracle does not support substring_index(). That is a MySQL function.
You can use regexp_substr(). Without sample data it is a little hard to be 100% sure, but I think the logic you want is:
regexp_substr(t3.fave_product, '[^:]+$') as fave_product
In Postgresql database I have a column called names where I have some names which need to be parsed using regex to clean up punctuation parts. I am able to get a clean name using regexp_replace as follows:
select regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g')
from tableA
However, I would like to compare with some strings that are also cleaned of punctuation. How can I use similar to with the formed regular expression?
select name
from tableA
where (lower(name) ~ '\.COM|''[A-Za-z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)') as nameParsed similar to '(fg )%' and
(lower(name) ~ '\.COM|''[A-Za-z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)') as nameParsed similar to '%( cargo| carrier| cartage )%'
With the previous query I am getting this error:
LINE 3: ...-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)') as namePar...
I have tried in where clause like this and it seems to be working:
select name
from tableA
where (select lower(regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g'))) similar to '(fg )%'
Is this the best approach? The execution time went to 46 seconds :(
Thanks in advance
You're trying to get a column name in a WHERE clause (is a comparison, not a column). So, you can use as follows:
SELECT name
FROM "tableA"
WHERE (regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g') similar to '(fg )%'
OR regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g') similar to '%( cargo| carrier| cartage )%');
Alternatively, you can use ilike instead of similar to if you want to find a specific word.
I would like to understand why this does not work :
select my_table.my_column AS 'what I want to write'
from my_table
The error:
ORA-00923: FROM keyword not found where expected
How do I name my column what I want to write ?
Thank you
Single quotes are for string literals, double quotes are for delimited identifiers, e.g. "what I want to write".
(sorry for my poor english)
If you try this select operation over a sqlite database:
SELECT column AS 'alias 1' FROM table;
You get the expected column name:
alias 1
--------
result 1
result 2
but if your alias contains a dot "." ... you get a wrong column name:
SELECT column AS 'alias.1' FROM table;
1
--------
result 1
result 2
(all behind the dot is ommited in the column name)
Wow...
It's weird...
anyone can help me?
thank you very much
UPDATE:
maybe it's just a bug in SQLiteStudio (the software where I'm testing my queries) and in QT (they both doesn't expect dots in alias names but sqlite does)
Enclose your alias in double quotes.
SELECT 'test' AS "testing.this"
Output:
| testing.this |
test
Updated:
Double quotes are used to enclose identifiers in SQL, not single quotes. Single quotes are only for strings. In this case you are trying to ensure that "testing.this" is used as is and not confused as testing.this (testing table this column).
http://www.sqlite.org/faq.html#q24
Use backticks
SELECT column AS `alias.1` FROM table;
Or double quotes (ANSI standard) per the other answer
SELECT column AS "alias.1" FROM table;
Both verified in SQLite Manager for FireFox
Definitely working properly:
C:\Windows>sqlite3.exe
SQLite version 3.7.8 2011-09-19 14:49:19
Enter ".help" for instructions
Enter SQL statements terminated with a ";"
sqlite> .mode column
sqlite> .headers on
sqlite> SELECT 'hello' AS 'alias.1';
alias.1
----------
hello
sqlite>
If you're using the SQLite 3 then the following query works just fine with various types used for the Alias column names.
See the result below the query:
select '1' as 'my.Col1', '2' as "my.Col2", '3' as [my.Col3], '4' as [my Col4] , '5' as 'my Col5'
I've found a "fix"...
SELECT column AS '.alias.1' FROM
table;
alias.1
--------
result 1
result 2
just another dot in the begining...
of course I don't like this solution...
any other idea??
Please try below, it works on Hive
select 1 as `xxx.namewith.dot`
xxx. means any word you want to input with dot notation on latest
namewith.dot means any alias name with dot notation on it