SQL中标记系统的布尔表达式

时间:2009-12-02 00:52:37

标签: sql boolean tagging

将此SQL表用于标记系统:

CREATE TABLE tags (
    id SERIAL PRIMARY KEY,
    name VARCHAR(100)
);
CREATE INDEX tags_name_idx ON tags(name);

CREATE TABLE tagged_items (
    tag_id INT,
    item_id INT
);
CREATE INDEX tagged_items_tag_id_idx ON tagged_items(tag_id);
CREATE INDEX tagged_items_item_id_idx ON tagged_items(item_id);

CREATE TABLE items (
    id SERIAL PRIMARY KEY,
    content VARCHAR(255)
);

SQL中用户的布尔表达式查询“ tag1 AND tag2 ”是:

SELECT items.* FROM items
    INNER JOIN tagged_items AS i1 ON (items.id = i1.item_id) INNER JOIN tags AS t1 ON (i1.tag_id = t1.id)
    INNER JOIN tagged_items AS i2 ON (items.id = i2.item_id) INNER JOIN tags AS t2 ON (i2.tag_id = t2.id)
WHERE t1.name = 'tag1' AND t2.name = 'tag2';

如何使用布尔表达式翻译其他查询,例如“ tag1 OR tag2 AND tag3 ”......或更复杂的查询,例如“ tag1 AND(tag2 OR tag3) AND NOT tag4 OR tag5 “到SQL?

1 个答案:

答案 0 :(得分:2)

假设数据 - >项目,单词 - > name和tagged_item - > tagged_items。

这是针对“ tag1 AND(tag2 OR tag3)AND NOT tag4 OR tag5 ”。我相信你可以弄明白其余的。

SELECT items.* FROM items
    LEFT JOIN (SELECT i1.item_id FROM tagged_items AS i1 INNER JOIN tags AS t1 ON i1.tag_id = t1.id AND t1.name = 'tag1') AS ti1 ON items.id = ti1.item_id
    LEFT JOIN (SELECT i2.item_id FROM tagged_items AS i2 INNER JOIN tags AS t2 ON i2.tag_id = t2.id AND t2.name = 'tag2') AS ti2 ON items.id = ti2.item_id
    LEFT JOIN (SELECT i3.item_id FROM tagged_items AS i3 INNER JOIN tags AS t3 ON i3.tag_id = t3.id AND t3.name = 'tag3') AS ti3 ON items.id = ti3.item_id
    LEFT JOIN (SELECT i4.item_id FROM tagged_items AS i4 INNER JOIN tags AS t4 ON i4.tag_id = t4.id AND t4.name = 'tag4') AS ti4 ON items.id = ti4.item_id
    LEFT JOIN (SELECT i5.item_id FROM tagged_items AS i5 INNER JOIN tags AS t5 ON i5.tag_id = t5.id AND t5.name = 'tag5') AS ti5 ON items.id = ti5.item_id
WHERE ti1.item_id IS NOT NULL AND (ti2.item_id IS NOT NULL OR ti3.item_id IS NOT NULL) AND ti4.item_id IS NULL OR ti5.item_id IS NOT NULL;

修改 如果要避免子查询,可以这样做:

SELECT items.* FROM items 
    LEFT JOIN tagged_items AS i1 ON items.id = i1.item_id LEFT JOIN tags AS t1 ON i1.tag_id = t1.id AND t1.name = 'tag1'
    ...
WHERE t1.item_id IS NOT NULL ...

我不确定你为什么要这样做,因为额外的左连接可能会导致运行速度变慢。