Question

我需要写一个这样的查询：

SELECT * FROM users WHERE user_name='somthing' ELSE WHERE user_name LIKE 'something%' ORDER BY user_name;

但我不确定如何写它。

表格如下：

id | user_name | frequency

第一条记录是

4 | fred | 1

所以当搜索user_name =＆＃39; fred＆＃39;如果有完全匹配，则显示最常见的记录（最高频率值）否则如果没有完全匹配则显示结果，例如＆＃39; fred＆＃39;

我复制了一个例子。链接是http://www.peachpit.com/articles/article.aspx?p=1802754&seqNum=3

Answer 1

CREATE TABLE users
(
    id INT,
    user_name VARCHAR(255),
    frequency INT
);

INSERT INTO users
VALUES
(4,'fred',1),
(5,'fred',2),
(6,'frederick',1),
(7,'freddie',1),
(8,'freddie',2),
(9,'john',1);

(
    SELECT  *
    FROM    users
    WHERE   user_name = 'fred'
    ORDER BY Frequency DESC
    LIMIT   1
)

UNION

(
    SELECT  *
    FROM    users
    WHERE   user_name LIKE '%fred%' AND
            NOT EXISTS (
                            SELECT  *
                            FROM    users
                            WHERE   user_name = 'fred'
                            ORDER BY Frequency
                            LIMIT   1
                        )
)

Mysql where =＆＃39; something＆＃39;用于搜索

1 个答案: