在此系统中,我们存储产品,产品图像(产品可能有许多图像),以及产品的默认图像.数据库:
CREATE TABLE `products` (
`ID` int(10) unsigned NOT NULL AUTO_INCREMENT,
`NAME` varchar(255) NOT NULL,
`DESCRIPTION` text NOT NULL,
`ENABLED` tinyint(1) NOT NULL DEFAULT '1',
`DATEADDED` datetime NOT NULL,
`DEFAULT_PICTURE_ID` int(10) unsigned DEFAULT NULL,
PRIMARY KEY (`ID`),
KEY `Index_2` (`DATEADDED`),
KEY `FK_products_1` (`DEFAULT_PICTURE_ID`),
CONSTRAINT `FK_products_1` FOREIGN KEY (`DEFAULT_PICTURE_ID`) REFERENCES `products_pictures` (`ID`) ON DELETE SET NULL ON UPDATE SET NULL
) ENGINE=InnoDB AUTO_INCREMENT=30 DEFAULT CHARSET=utf8;
CREATE TABLE `products_pictures` (
`ID` int(10) unsigned NOT NULL AUTO_INCREMENT,
`IMG_PATH` varchar(255) NOT NULL,
`PRODUCT_ID` int(10) …Run Code Online (Sandbox Code Playgroud) 关于这个答案,我偶然发现了一个我无法解释的现象.
版本:
x86_64-unknown-linux-gnu上的PostgreSQL 9.1.2,由gcc-4.4.real编译(Debian 4.4.5-8)4.4.5,64位
请考虑以下演示.测试平台:
CREATE TEMP TABLE t (
id integer
,txt text
,CONSTRAINT t_pkey PRIMARY KEY (id) DEFERRABLE INITIALLY IMMEDIATE
);
INSERT INTO t VALUES
(1, 'one')
,(2, 'two');
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UPDATE t
SET id = t_old.id
FROM t t_old
WHERE (t.id, t_old.id) IN ((1,2), (2,1));
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目前的实施似乎有一个错误?上面的UPDATE虽然不应该有效.约束已定义INITIALLY IMMEDIATE,我没有使用SET CONSTRAINTS.
我错过了什么,或者这是一个(相当无害)的错误?
因此,修改CTE的数据也可以工作,尽管它有一个NOT DEFERREDpk 失败:
WITH x AS (
UPDATE t SET id = 1 WHERE id = 2 …Run Code Online (Sandbox Code Playgroud) 我有一个定义相互递归表的模型:
Answer
questionId QuestionId
text
Question
text
correct AnswerId
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实际插入问题需要做什么?我需要先知道正确答案是什么.但要插入答案,我需要知道它回答了什么问题.
如果重要的话,我正在运行Postgres.
DDL是:
CREATE TABLE answer (
id integer NOT NULL, -- answer id
text character varying NOT NULL, -- answer text
question_id bigint NOT NULL -- question id
);
CREATE TABLE question (
id integer NOT NULL, -- question id
question character varying NOT NULL, -- question text
correct bigint NOT NULL, -- correct answer
solution character varying NOT NULL -- solution text
);
ALTER TABLE ONLY answer ALTER COLUMN …Run Code Online (Sandbox Code Playgroud) sql postgresql database-design referential-integrity foreign-keys