我有一个包含数字列表的 SQLite 表。如下所示:
1
2
3
8
9
11
12
13
14
15
18
19
我需要将它们变成具有顺序 ID 的块。例如,
1-3
8-9
11-15
18-19
这可能只使用SQL,还是我需要循环并比较以前的和当前的......
我找到了使用临时表的解决方案。
首先,我们必须确定起点和终点:
http://sqlfiddle.com/#!5/fdc26/13
SELECT
a.x AS ax,
CASE WHEN p.x IS NULL THEN 1
WHEN n.x IS NULL THEN 2
ELSE 0
END AS begin_or_end
FROM num AS a
LEFT JOIN num AS n ON a.x + 1 = n.x /* n: next */
LEFT JOIN num AS p ON a.x - 1 = p.x /* p: prev */
WHERE p.x IS NULL
OR n.x IS NULL
ORDER BY a.x ASC;
或者,您可以将其拆分为两个查询:
SELECT a.x AS begin_point
FROM num AS a
LEFT JOIN num AS p ON a.x - 1 = p.x /* p: prev */
WHERE p.x IS NULL
ORDER BY a.x ASC;
SELECT a.x AS end_point
FROM num AS a
LEFT JOIN num AS n ON a.x + 1 = n.x /* n: next */
WHERE n.x IS NULL
ORDER BY a.x ASC;
我选择了第一个选项来创建lohi
在我的示例中调用的临时表。我使用每个表都有一个自动的事实rowid
(有关更多详细信息,请参见http://www.sqlite.org/lang_createtable.html/ROWIDs和 INTEGER PRIMARY KEY)
最终查询:
http://sqlfiddle.com/#!5/21770/1
SELECT lo.ax, hi.ax
FROM lohi AS lo
INNER JOIN lohi AS hi ON lo.rowid+1 = hi.rowid
WHERE lo.rowid % 2;