Question

您好我正在尝试在mysql视图中循环实现行自动编号。我确实从Pure-SQL Technique for Auto-Numbering Rows in Result Set开始并让它发挥作用。然后我尝试调整它并在那里得到ALMOST ......现在我是stuMainmped

该视图列出了狗（CaneID）及其后代（AvoID）。我想要实现的是每次主狗（CaneID）改变时重新编号，这样我就可以让每个上升者属于一个从1开始的狗号。

到目前为止它仅适用于第一只狗，以下仍按顺序编号但不从1开始。

这是我的查询查询：

SELECT
`avi1`.`CaneID`,
`avi1`.`Cane`,
`avi1`.`PedPos`,
`avi1`.`AvoID`,
`avi1`.`Avo`,
`avi1`.`Ripetuto`,
(
    SELECT
        count(0)
    FROM
        `v_avi_ripetuti` `Avi2`
    WHERE
        (
             (`avi2`.`Ripetuto` > 1)
            AND (`avi2`.`CaneID` <= `avi1`.`CaneID`)
            AND (`avi2`.`AvoID` <= `avi1`.`AvoID`)

        )
    ORDER BY
        `avi2`.`CaneID`,
        `avi2`.`AvoID`
) AS `RowNumber`
FROM
`v_avi_ripetuti` `Avi1`
WHERE
(`avi1`.`Ripetuto` > 1)
ORDER BY
`avi1`.`CaneID`,
`avi1`.`AvoID`

我现在得到的是：

CaneID  Cane         AvoID  Avo                             RowNumber
---------------------------------------------------------------------
2       Antigua      472    Anika v. Stammhaus Eike         1
2       Antigua      527    Britta v.d. Römerlinde          2
2       Antigua      642    Ares v. Nettenberg              3
2       Antigua      657    Carmen v. Westfalenzwinger      4
2       Antigua      658    Leu v. Stammhaus Eike           5
2       Antigua      684    Chipsy v. Theresienhof          6
2       Antigua      1662   Astrit v.d. Burrlinde           7
-----------------------> New Main dog<-------------------------------
3       Ambra Gaya   472    Anika v. Stammhaus Eike         2
3       Ambra Gaya   657    Carmen v. Westfalenzwinger      6
3       Ambra Gaya   658    Leu v. Stammhaus Eike           8
3       Ambra Gaya   684    Chipsy v. Theresienhof          10
3       Ambra Gaya   718    Leo v. guten Löwen              11
3       Ambra Gaya   1022   Zara v. Ratibor u. Corvey       12
3       Ambra Gaya   1024   Ruth v. Ratibor u. Corvey       13
3       Ambra Gaya   1050   Dunja v. Hofoldinger Forst      14
3       Ambra Gaya   1200   Icksle v. Ratibor u. Corvey     15
3       Ambra Gaya   1227   Troldegaards Tanja              16
3       Ambra Gaya   1228   Anemone v. Ratibor u. Corvey    17
-----------------------> New Main dog<-------------------------------
3371    Always Habit 600    Absinthe v. Kallenberg          4
3371    Always Habit 750    Mathoaka's Relaxing Boy         13
3371    Always Habit 2560   Leijonamielen Luvaton Lyyli     21

我是mysql的新手（一周前安装它）并且非常努力地让访问权限落后......请耐心等待。

Answer 1

我猜你想要按照结果排序，首先是CaneId，然后是AvoId。如果是这样，正确的子查询是：

(SELECT count(*)
 FROM v_avi_ripetuti` `Avi2`
  WHERE `avi2`.`Ripetuto` > 1 AND
        (`avi2`.`CaneID` < `avi1`.`CaneID` or
          `avi2`.`CaneID` = `avi1`.`CaneID` and avi2`.`AvoID` <= `avi1`.`AvoID`
        )
) AS `RowNumber`

除了修复where子句之外，我还删除了order by，因为它与子查询无关。我还将count(0)更改为count(*)。这些是等价的，但count(0)看起来很尴尬。

Answer 2

我知道!!!!!!! 我使用函数SUM（）而不是COUNT（）。这是我的mysql查看代码，万一其他人可以使用它：

SELECT
`avi1`.`CaneID`,
`avi1`.`Cane`,
`avi1`.`PedPos`,
`avi1`.`AvoID`,
`avi1`.`Avo`,
`avi1`.`Ripetuto`,
(
    SELECT
        sum(

            IF (
                (
                    (
                        `avi2`.`CaneID` = `avi1`.`CaneID`
                    )
                    AND (`avi2`.`AvoID` <= `avi1`.`AvoID`)
                ),
                1,
                0
            )
        )
    FROM
        `v_avi_ripetuti` `Avi2`
    WHERE
        (`avi2`.`Ripetuto` > 1)
) AS `NumRiga`
  FROM
`v_avi_ripetuti` `Avi1`
 WHERE
(`avi1`.`Ripetuto` > 1)

这给了我想要的结果：

 CaneID  Cane         AvoID  Avo                             RowNumber
 ---------------------------------------------------------------------
 2       Antigua      472    Anika v. Stammhaus Eike         1
 2       Antigua      527    Britta v.d. Römerlinde          2
 2       Antigua      642    Ares v. Nettenberg              3
 2       Antigua      657    Carmen v. Westfalenzwinger      4
 2       Antigua      658    Leu v. Stammhaus Eike           5
 2       Antigua      684    Chipsy v. Theresienhof          6
 2       Antigua      1662   Astrit v.d. Burrlinde           7
 -----------------------> New Main dog<-------------------------------
 3       Ambra Gaya   472    Anika v. Stammhaus Eike         1
 3       Ambra Gaya   657    Carmen v. Westfalenzwinger      2
 3       Ambra Gaya   658    Leu v. Stammhaus Eike           3
 3       Ambra Gaya   684    Chipsy v. Theresienhof          4
 3       Ambra Gaya   718    Leo v. guten Löwen              5
 3       Ambra Gaya   1022   Zara v. Ratibor u. Corvey       6
 3       Ambra Gaya   1024   Ruth v. Ratibor u. Corvey       7
 3       Ambra Gaya   1050   Dunja v. Hofoldinger Forst      8
 3       Ambra Gaya   1200   Icksle v. Ratibor u. Corvey     9
 3       Ambra Gaya   1227   Troldegaards Tanja              10
 3       Ambra Gaya   1228   Anemone v. Ratibor u. Corvey    11
 -----------------------> New Main dog<-------------------------------
 3371    Always Habit 600    Absinthe v. Kallenberg          1
 3371    Always Habit 750    Mathoaka's Relaxing Boy         2
 3371    Always Habit 2560   Leijonamielen Luvaton Lyyli     3

在循环中以mysql视图自动编号行

2 个答案: