3 Table JOIN to show everything – Mysql
I want a query from my 3 table, but i dont know which JOIN should I use. I want to display all of table2's item_id and item_name, also show stat and shipped from table3 even if its NULL. Also, user_id…
I want a query from my 3 table, but i dont know which JOIN should I use. I want to display all of table2's item_id and item_name, also show stat and shipped from table3 even if its NULL. Also, user_id…
i have solved 99% challenges in sql in skillrack.com which is a site for practicing coding for placements, this is the last one, I cant even understand the question itself, I think there is a conceptual error in question, If…
I have an array['a', 'b', 'c'] Very simply I want to check whether all elements of the above array exist in a column say test of a table t. It would be best if the sql return boolean. Thanks.
Using this table: CREATE TABLE t (i int PRIMARY KEY, j variant); I'm trying to insert some JSON object into the table, e.g. INSERT INTO t (i, j) VALUES (1, object_construct('a', 1)); However, that fails with: SQL Error [2014] [22000]:…
I want to only display searched string from a table, as example this is my table: Table name: guidelines id content 1 An individual is accused “of” a crime, not “with” or “for” a crime. Accused, often as “the accused”,…
There exists the following table: practice=# select * from table; letter | value | year --------+---------+------ A | 5000.00 | 2021 B | 6000.00 | 2021 C | 6000.00 | 2021 B | 8000.00 | 2022 A | 9000.00 |…
I am using MySql and running the following queries. (select CONCAT(RESOURCE, " AMB:", AMB) AS REC_1 from db.CRE where CONTENT = '183' and LEVEL = '99' EXCEPT select CONCAT(RESOURCE, " AMB:", AMB) AS REC_1 from db.CRE where CONTENT = '182'…
I installed the clickhouse program in the ubuntu operating system and connected the SQLyog program. I can create a database, but I cannot create a table in it. It gives the following CODE:119 error. ubuntu :) create table taxonomy_object_firewalls CREATE…
In postgressql is there a way to attain the result below by using partition by or any other way? last_name year increment partition Doe 2000 1 1 Doe 2001 2 1 Doe 2002 3 1 Doe 2003 -1 2 Doe…
I need to workout the effectiveness of a user to display in a report, i complied the following statement SELECT COUNT(`call_id`) AS logged FROM `tbl_calls` WHERE `user_id_attended_by` = 24 AND YEAR(`date_ack_by_tech`) = YEAR(CURRENT_DATE - INTERVAL 1 YEAR) AND MONTH(`date_ack_by_tech`) =…