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      2. MySQL:选择语句中的自动增量临时列

        MySQL: Auto increment temporary column in select statement(MySQL:选择语句中的自动增量临时列)

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                  本文介绍了MySQL:选择语句中的自动增量临时列的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  如何在 MySQL 的 select 语句中创建和自动增加一个临时列?

                  这是我目前所拥有的:

                  SET @cnt = 0;选择(@cnt =@cnt + 1) AS rowNumber,行号从我的表类别 ID = 1

                  返回:

                  +++++++++++++++++++++++ 行号 |行号 +++++++++++++++++++++++++ (NULL) |1 ++ (NULL) |25 ++ (NULL) |33 ++ (NULL) |150++ (NULL) |219 ++++++++++++++++++++++++

                  但我需要:

                  +++++++++++++++++++++++ 行号 |行号 +++++++++++++++++++++++++ 1 |1 ++ 2 |25 ++ 3 |33 ++ 4 |150++ ... |... ++++++++++++++++++++++++

                  解决方案

                  这会给你一个连续行号和3.

                  SELECT(@cnt := @cnt + 1) AS rowNumber,行号从 myTable AS tCROSS JOIN (SELECT @cnt := 0) AS 虚拟哪里 t.CategoryID = 1按 t.rowID 排序;

                  结果

                  <前>|行号 |行列 |---------------------|1 |1 ||2 |25 ||3 |33 ||4 |150 |

                  How do I create and auto increment a temporary column in my select statement with MySQL?

                  Here is what I have so far:

                  SET @cnt = 0;
                  SELECT
                      (@cnt =@cnt + 1) AS rowNumber,
                      rowID
                  FROM myTable
                  WHERE CategoryID = 1
                  

                  Which returns:

                  +++++++++++++++++++++
                  + rowNumber | rowID +
                  +++++++++++++++++++++
                  +  (NULL)   |   1   +
                  +  (NULL)   |   25  +
                  +  (NULL)   |   33  +
                  +  (NULL)   |   150 +
                  +  (NULL)   |   219 +
                  +++++++++++++++++++++
                  

                  But I need:

                  +++++++++++++++++++++
                  + rowNumber | rowID +
                  +++++++++++++++++++++
                  +  1        |   1   +
                  +  2        |   25  +
                  +  3        |   33  +
                  +  4        |   150 +
                  +  ...      |   ... +
                  +++++++++++++++++++++
                  

                  解决方案

                  This will give you a consecutive row number with 3.

                  SELECT
                      (@cnt := @cnt + 1) AS rowNumber,
                      t.rowID
                  FROM myTable AS t
                    CROSS JOIN (SELECT @cnt := 0) AS dummy
                  WHERE t.CategoryID = 1
                  ORDER BY t.rowID ;
                  

                  Result

                  | ROWNUMBER | ROWID |
                  ---------------------
                  |         1 |     1 |
                  |         2 |    25 |
                  |         3 |    33 |
                  |         4 |   150 |

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