Question

我有桌子

table1
****************************
id | start_date | end_date  
1    30/06/2012   01/01/9999

当我插入新行时

id | start_date | end_date
1    30/06/2012   26/06/2012 <- Now it is the start date of the next row2
2    26/06/2012   01/01/9999

插入行的开始日期将成为上一行的结束日期，依此类推。

 $sql = "INSERT INTO 
                    table1(cate_id,task_id,obj,end_date,start_date,line)
                    SELECT 
                        $cate_id ,
                        $task_id,
                        $obj,
                        '2011-02-28', -> this what I am trying to do
                        '9999-01-01', -> this what I am trying to do
                        line1 
                    FROM 
                        line_task
                    WHERE 
                         line_task_id = (
                                            SELECT 
                                                line_task_id 
                                            FROM 
                                                task 
                                            WHERE 
                                                task_id = 2
                                           )";

当我尝试这样的事情时：

insert into table1(cate_id,task_id,obj, start_date, end_date,line) 
values (4,3,23, (Select max(end_date) from table1),end_date,'value_line');

我收到了消息：

  Error code 1093, SQL state HY000: You can't specify target table 'table1' for update     in FROM clause
Line 1, column 1

执行完成0秒后，发生1个错误

可以告诉我怎样才能用mysql做到这一点？感谢

Answer 1

试试这个::

  Set @max_date = Select max(end_date) from table;

    Insert into table(id, start_date, end_date) values (id, @max_date ,end_date);

Answer 2

首先，获取最后一条记录end_date：

$sql = mysql_query("Select * from table1 ORDER BY ID Desc");

    if ($row=mysql_fetch_array($sql))
    {
        $end_date=$row['end_date'];
    }

然后使用之前的$ end_date插入新记录。

Answer 3

您可以在一个查询中执行此操作

INSERT INTO table1(cate_id,task_id,obj, start_date, end_date,line) 
SELECT 4,3,23, MAX(end_date), end_date, 'value_line'
FROM table1;

只需遵循正确的语法（第二个示例）：http://dev.mysql.com/doc/refman/5.5/en/insert-select.html

在同一个表中插入和更新

3 个答案: