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1 lut 2010 · Here's a query to update a table based on a comparison of another table. If record is not found in tableB, it will update the "active" value to "n". If it's found, will set the value to NULL. UPDATE tableA LEFT JOIN tableB ON tableA.id = tableB.id SET active = IF(tableB.id IS NULL, 'n', NULL)"; Hope this helps someone else.
In this article, we would like to show you UPDATE query with IF condition in MySQL. Quick solution: UPDATE `table_name` SET `column_name` = IF(condition , if_true, if_false); Practical example. To show UPDATE query with IF condition, we will use the following users table:
The MySQL UPDATE Statement. The UPDATE statement is used to modify the existing records in a table. UPDATE Syntax. UPDATE table_name. SET column1 = value1, column2 = value2, ... WHERE condition; Note: Be careful when updating records in a table! Notice the . WHERE clause in the UPDATE statement.
21 sie 2024 · MySQL offers conditional control flow with IF statements, allowing us to execute blocks of SQL code depending on whether a condition is true or false. In this article, We will learn about What is Decision-Making in MySQL with different statements along with examples and so on.
Definition and Usage. The IF () function returns a value if a condition is TRUE, or another value if a condition is FALSE. Syntax. IF (condition, value_if_true, value_if_false) Parameter Values. Technical Details. More Examples. Example. Return 5 if the condition is TRUE, or 10 if the condition is FALSE: SELECT IF(500<1000, 5, 10);
DELIMITER // CREATE FUNCTION VerboseCompare (n INT, m INT) RETURNS VARCHAR(50) BEGIN DECLARE s VARCHAR(50); IF n = m THEN SET s = 'equals'; ELSE IF n > m THEN SET s = 'greater'; ELSE SET s = 'less'; END IF; SET s = CONCAT('is ', s, ' than'); END IF; SET s = CONCAT(n, ' ', s, ' ', m, '.');
16 kwi 2016 · UPDATE Tests AS old INNER JOIN ( SELECT 10 AS TestId, 25 AS TestSubId, 1000 AS TestScore, 2000 AS TestScore2 UNION ALL SELECT 11, 22, 1100, 2100 ) AS new ON old.TestId = new.TestId AND old.TestSubId = new.TestSubId SET old.TestScore = new.TestScore, old.TestScore2 = new.TestScore2 ;
