隐藏重复行sql phpmyadmin

Question

我有关于重复行的问题。

我的目的是这样的：

你能帮我解决一下这样的查询吗？

我一直使用的代码如下：

select a.id_task1, b.id_task, b.task, b.cycle_sequence_number, c.cycle_sequence_number FROM task_table a LEFT OUTER JOIN billing_cycle b ON (a.id_task1=b.id_task) LEFT OUTER JOIN header c ON (b.cycle_sequence_number = c.cycle_sequence_number) where a.id_task1< 23 order by b.cycle_sequence_number DESC;

此致 - 三 -

抱歉，先生，我忘了发布我的代码：

select a.id_task1, b.id_task, b.task, b.cycle_sequence_number, c.cycle_sequence_number
FROM task_table a 
    LEFT OUTER JOIN billing_cycle b ON (a.id_task1=b.id_task) 
    LEFT OUTER JOIN header c ON (b.cycle_sequence_number = c.cycle_sequence_number) 
where a.id_task1< 23
order by b.cycle_sequence_number DESC;

Answer 1

在关系数据库中，查询的结果是关系/表（'relation'和'table'是同义表达式）。因此，让我们为您的查询命名并使用group by s：

select id_task1, id_task, task, max(b.cycle_sequence_number) as b_cycle_sequence_number, max(c.cycle_sequence_number) as c_cycle_sequence_number
from (...) t
group by id_task1, id_task, task
order by b_cycle_sequence_number

您应将b.cycle_sequence_number解释为b_cycle_sequence_number，将c.cycle_sequence_number解释为c_cycle_sequence_number，以避免重复的列名称。很抱歉没有重写您的实际代码，但我拒绝从图片中输入您的代码。

编辑：真正的代码是：

select a.id_task1, b.id_task, b.task, b.cycle_sequence_number, c.cycle_sequence_number FROM task_table a LEFT OUTER JOIN billing_cycle b ON (a.id_task1=b.id_task) LEFT OUTER JOIN header c ON (b.cycle_sequence_number = c.cycle_sequence_number) where a.id_task1< 23 order by b.cycle_sequence_number DESC;

建议：

SELECT id_task1, id_task, task, max(b.cycle_sequence_number) as b_cycle_sequence_number, max(c.cycle_sequence_number) as c_cycle_sequence_number
FROM (SELECT a.id_task1, b.id_task, b.task, b.cycle_sequence_number as b_cycle_sequence_number, c.cycle_sequence_number as c_cycle_sequence_number
      FROM task_table a 
      LEFT OUTER JOIN billing_cycle b 
      ON (a.id_task1=b.id_task) 
      LEFT OUTER JOIN header c 
      ON (b.cycle_sequence_number = c.cycle_sequence_number) where a.id_task1< 23) t
GROUP BY id_task1, id_task, task
ORDER BY b_cycle_sequence_number