DISTINCT के साथ YEAR() और MONTH() (या DATE()) का उपयोग करके DATETIME कॉलम से अद्वितीय दिनांक प्राप्त करें।
एक टेबल बनाते हैं:
CREATE TABLE PostMessageDemo (
UserId INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
UserMessage VARCHAR(100),
UserPost DATETIME
);
रिकॉर्ड डालें:
INSERT INTO PostMessageDemo (UserMessage, UserPost) VALUES ('Software Developer', NOW());
INSERT INTO PostMessageDemo (UserMessage, UserPost) VALUES ('Software Developer', DATE_ADD(NOW(), INTERVAL 3 MONTH));
INSERT INTO PostMessageDemo (UserMessage, UserPost) VALUES ('Software Developer', NOW());
INSERT INTO PostMessageDemo (UserMessage, UserPost) VALUES ('Software Developer', '2019-03-31 17:30:34');
सभी रिकॉर्ड देखें:
SELECT * FROM PostMessageDemo;
आउटपुट
+--------+--------------------+---------------------+ | UserId | UserMessage | UserPost | +--------+--------------------+---------------------+ | 1 | Software Developer | 2019-02-15 18:07:27 | | 2 | Software Developer | 2019-05-15 18:09:18 | | 3 | Software Developer | 2019-02-15 18:09:31 | | 4 | Software Developer | 2019-03-31 17:30:34 | +--------+--------------------+---------------------+
विशिष्ट वर्ष-महीना संयोजन:
SELECT DISTINCT YEAR(UserPost), MONTH(UserPost) FROM PostMessageDemo;
आउटपुट
+----------------+-----------------+ | YEAR(UserPost) | MONTH(UserPost) | +----------------+-----------------+ | 2019 | 2 | | 2019 | 5 | | 2019 | 3 | +----------------+-----------------+
बेहतर तरीका (पूर्ण दिनांक):
SELECT DISTINCT DATE(UserPost) FROM PostMessageDemo;
यह 2019-02-15, 2019-05-15, 2019-03-31 लौटाएगा।