如何執行MySQL UPSERT

這就是我的表的樣子。如何執行MySQL UPSERT

CREATE TABLE `answers` (
`id` INT(11) NOT NULL AUTO_INCREMENT, 
`basicinfo_id` INT(11) NOT NULL, 
`question_id` INT(11) NOT NULL, 
`answer` INT(11) NOT NULL, 
PRIMARY KEY (`id`), 
INDEX `question_id` (`question_id`), 
INDEX `basicinfo_id` (`basicinfo_id`), 
CONSTRAINT `basicinfo_id` FOREIGN KEY (`basicinfo_id`) REFERENCES `basic_info` (`id`) ON UPDATE NO ACTION ON DELETE NO ACTION, 
CONSTRAINT `question_id` FOREIGN KEY (`question_id`) REFERENCES `questions` (`id`) ON UPDATE NO ACTION ON DELETE NO ACTION) COLLATE='latin1_swedish_ci' ENGINE=InnoDB AUTO_INCREMENT=3383;

這是數據如何能像

我想，如果他們已經存在「basicinfo_id」更新的答案。或如果我更改question_id 1，2和3的答案。我該如何插入數據。

我試過這個查詢，但它沒有更新結果。

INSERT INTO answers (basicinfo_id, question_id, answer) VALUES('98', 1, '1'),('98', 2, '1'),('98', 3, '1'),('98', 4, '1'),('98', 5, '1') ON DUPLICATE KEY UPDATE basicinfo_id = 98;

我已經定義了一個複合唯一鍵現在這是它的樣子，但它仍然沒有工作。

CREATE TABLE `answers` (
`id` INT(11) NOT NULL AUTO_INCREMENT, 
`basicinfo_id` INT(11) NOT NULL, 
`question_id` INT(11) NOT NULL, 
`answer` INT(11) NOT NULL, 
PRIMARY KEY (`id`), 
UNIQUE INDEX `basicinfo_id_question_id` (`basicinfo_id`, `question_id`), 
INDEX `question_id` (`question_id`), 
INDEX `basicinfo_id` (`basicinfo_id`), 
CONSTRAINT `basicinfo_id` FOREIGN KEY (`basicinfo_id`) REFERENCES `basic_info` (`id`) ON UPDATE NO ACTION ON DELETE NO ACTION, 
CONSTRAINT `question_id` FOREIGN KEY (`question_id`) REFERENCES `questions` (`id`) ON UPDATE NO ACTION ON DELETE NO ACTION) COLLATE='latin1_swedish_ci' ENGINE=InnoDB AUTO_INCREMENT=3718;

來源

2016-08-23 Shaonline

永遠不會發生更新插入。你永遠不會導致重複的密鑰違規。只有當存在唯一的密鑰違規時，「重複密鑰」纔會實際觸發。沒有重點違規，沒有更新。 –

對不起，如果我來這裏有點無知，但如果我設置basicinfo_id爲UNIQUE它不會允許重複basicinfo_id，我不想但如果我設置basicinfo_id和question_id都作爲複合獨特的關鍵，那麼它是可能的。我有什麼意義嗎？ – Shaonline

是的，在這種情況下，您需要一個複合唯一密鑰。 –

你ON DUPLICATE KEY UPDATE basicinfo_id = 98;是畸形

試試這個： ON DUPLICATE KEY UPDATE 'answer' = VALUES('answer');

如果一個唯一的密鑰已經存在，它會更新answer場

來源

2016-08-23 19:53:03 Sylwit

如何執行MySQL UPSERT

回答

相關問題