Make exact match show at top

SELECT * 
FROM client
WHERE identifyingnumber LIKE '%86%'
ORDER BY LEN(identifyingnumber)


SELECT *
FROM
(
SELECT *, 1 AS PRIO FROM client
WHERE identifyingnumber = '86'
UNION
SELECT *, 2 AS PRIO FROM client
WHERE identifyingnumber LIKE '%86%'
AND identifyingnumber <>'86'
) AS X
ORDER BY PRIO



SELECT *
FROM client
WHERE identifyingnumber LIKE '%86%'
ORDER BY 
    CASE WHEN identifyingnumber = '86' THEN 1 
        WHEN identifyingnumber LIKE '86%' THEN 2 ELSE 3 END


SELECT *, 
    CASE 
        WHEN identifyingnumber = '86' THEN 1
        WHEN identifyingnumber LIKE '86%' THEN 2
        ELSE 3
    END AS Rank
FROM client
WHERE identifyingnumber LIKE '%86%'
ORDER BY Rank



ORDER BY identifyingnumber = '86' DESC
posted @ 2015-03-11 15:13  罗森党  阅读(323)  评论(0编辑  收藏  举报