如何查看准备好的语句或者我应该放弃动态 table 名称？

Question

我正在努力更新几年前某人写的一些 PHP/MySQL 报告代码。单个 PHP 页面中有 50 多个 SQL 语句。其中一些创建临时tables，它们都引用至少一个动态table（例如Report_201701），这在PHP中使用变量很简单。

但是当我将所有这些语句调整为 MySQL Workbench 以进行故障排除和增强时，这会导致问题，因为我无法查看准备好的语句的输出，因为我正在使用动态 table 名称的变量。

我不熟悉动态 SQL。有什么方法可以 view/output 查询结果吗？ According to this post 看来我做不到。

几个语句 return 一个单一的值，我已经能够通过分配给这样的变量来查看：

set @sqlQuery = 
    Concat(
        'SELECT 1 as "A"
        into @sqlOut;'
    );
    #$opentickets = mysql_result($sqlQuery, 0);
PREPARE stmt1 from @sqlQuery;
EXECUTE stmt1;
deallocate prepare stmt1;
select @sqlOut;

# @sqlOut
-----------
1

但是如果我尝试使用 return 多列或多行的结果，我会得到如下错误：

set @sqlQuery = 
    Concat(
        'SELECT 1 as "A", 2 as "B", 3 as "C"
        into @sqlOut;'
    );
    #$opentickets = mysql_result($sqlQuery, 0);
PREPARE stmt1 from @sqlQuery;
EXECUTE stmt1;
deallocate prepare stmt1;
select @sqlOut;

Error Code: 1222. The used SELECT statements have a different number of columns


set @sqlQuery = 
    Concat(
        'SELECT 1 as "A"
        UNION SELECT 2 as "A"
        UNION SELECT 3 as "A"
        into @sqlOut;'
    );
    #$opentickets = mysql_result($sqlQuery, 0);
PREPARE stmt1 from @sqlQuery;
EXECUTE stmt1;
deallocate prepare stmt1;
select @sqlOut;

Error Code: 1172. Result consisted of more than one row

我最好的方法是什么？暂时用静态名称替换动态 table 名称？（我正在考虑永久这样做，因为在这种情况下 temp tables 不需要具有指示日期的名称。）

Answer 1

好吧，我偶然发现 MySQL Workbench 会输出准备好的 SELECT 语句。无需分配给变量。不确定这是否对某人有帮助，或者是否应该删除它以说明显而易见的问题。

set @sqlQuery = 
    Concat(
        'SELECT 1 as "A", 2 as "B", 3 as "C"
        UNION SELECT 10 as "A", 20 as "B", 30 as "C"
        UNION SELECT 100 as "A", 200 as "B", 300 as "C";'
    );
PREPARE stmt1 from @sqlQuery;
EXECUTE stmt1;
deallocate prepare stmt1;

Returns:

A       B       C
'1'     '2'     '3'
'10'    '20'    '30'
'100'   '200'   '300'

如何查看准备好的语句或者我应该放弃动态 table 名称？

How to view a prepared statement OR should I abandon dynamic table names?

mysql

dynamic-sql