MySQL的：选择行与前面和后面的行ID按日期

Question

我有这个表：

update_id | project_id | content | date 
------------------------------------------------------ 
1   | 1   | text... | 2011-12-20 22:10:30 
2   | 2   | text... | 2011-12-20 22:10:30 
3   | 2   | text... | 2011-12-21 22:10:30 
4   | 2   | text... | 2011-12-22 22:10:30 
5   | 2   | text... | 2011-12-23 22:10:30 

我当前的查询是：

SELECT update_id, project_id, content, date 
FROM update 
WHERE update_id = 2 
AND project_id = 2 
ORDER BY date DESC 
LIMIT 1 

但是，除了选定的行，我想也是基于日期列的前一行和下一行（对于特定的project_id），所以我可以显示'上一次更新'和'下一次更新'链接（如果可用）。

Answer 1

你需要2个子选择..但现在要链接它们。

SELECT @rownum:=@rownum+1 as place, update_id, project_id, content, date,() as previous,() as next, (SELECT @rownum:=0) 

其中blank（）被替换为相同形式的子查询。

Answer 2

第一种方法。这选择返回3行。您选择的行和前一个日期的行以及下一个日期的阳极行：

SELECT * FROM ( 
SELECT 
    'THIS' as update_time, 
    update_id, 
    project_id, 
    content, 
    date 
FROM 
    `update` 
WHERE 
    update_id = 2 AND project_id = 2 
ORDER 
    BY date DESC 
LIMIT 1 
) TT 
UNION 
SELECT * FROM ( 
SELECT 
    'PREVIOUS' as update_time, 
    update_id, 
    project_id, 
    content, 
    date 
FROM 
    `update` 
WHERE 
    update_id <> 2 AND project_id = 2 and 
    date <= (select date from `update` where update_id = 2) 
ORDER 
    BY date DESC 
LIMIT 1 
) TP 
UNION 
SELECT * FROM ( 
SELECT 
    'NEXT' as update_time, 
    update_id, 
    project_id, 
    content, 
    date 
FROM 
    `update` 
WHERE 
    update_id <> 2 AND project_id = 2 and 
    date >= (select date from `update` where update_id = 2) 
ORDER 
    BY date ASC 
LIMIT 1 
) TN