Question

我有一个包含数字列表的SQLite表。看起来如下：

我需要将这些转换为具有顺序ID的块。例如，

这是否可以仅使用SQL，或者我是否需要循环并比较以前的当前...

Answer 1

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;

我在我的示例中选择了第一个创建名为lohi的临时表的选项。我使用的事实是每个表都有一个自动rowid（有关详细信息，请参阅http://www.sqlite.org/lang_createtable.html / ROWID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;

在SQLite数据库中找到顺序id块

1 个答案: