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For a bit of completeness - this query will create a query to give you a count of all of the tables for a given owner. select DECODE(rownum, 1, '', ' UNION ALL ') || 'SELECT ''' || table_name || ''' AS TABLE_NAME, COUNT(*) ' || ' FROM ' || table_name as query_string from all_tables where owner = :owner; The output is something like
TRANSACT table who have done transaction less than or equal to 3000 (v) SELECT ANO, ANAME FROM ACCOUNT WHERE ADDRESS NOT IN ('CHENNAI', 'BANGALORE'); (vi) SELECT DISTINCT ANO FROM TRANSACT (vii) SELECT ANO, COUNT(*), MIN(AMOUNT) FROM TRANSACT GROUP BY ANO HAVING COUNT(*)> 1 (viii) SELECT COUNT(*), SUM(AMOUNT) FROM TRANSACT
COUNT(*) Code language: SQL (Structured Query Language) (sql) The COUNT(*) function returns the number of rows in a table in a query. It counts duplicate rows and rows that contain null values.
Here we use the COUNT() function and the GROUP BY clause, to return the number of records for each category in the Products table: Example SELECT COUNT(*) AS [Number of records], CategoryID
5 lut 2024 · These 15 MySQL exercises are especially designed for beginners. They cover essential topics like selecting data from one table, ordering and grouping data, and joining data from multiple tables. This article showcases beginner-level MySQL practice exercises, including solutions and comprehensive explanations. If you need to practice …
This 3-page SQL Cheat Sheet provides you with the most commonly used SQL statements. Download the SQL cheat sheet, print it out, and stick to your desk.
SELECT column_name_1, column_name_2 FROM table_name_1 INNER JOIN table_name_2 ON table_name_1.column_name_1 = table_name_2.column_name_1; Joining data in SQL: SQL Intermediate: Joins & Complex Queries Joining tables with INNER JOIN: SELECT * FROM facts LEFT JOIN cities ON cities.facts_id = facts.id; Joining tables using a LEFT JOIN: SELECT f ...
