php - @variable mysql 在 php mysql 执行时返回空值

Question

我需要在 php 中运行 mysql:

$sql_subject_summary = "SELECT 
                c.subject_code_id, c.subject_name, @total_target:= SUM(s.total_target_question) AS total_target_question,
                @total_correct := ROUND((RAND() * (@total_target-10))+10) AS total_correct, 
                (@total_correct / @total_target)*100 AS percent, c.icon_filename
                 FROM 
                edu_subject_code c LEFT JOIN wkp_wg_student_subject s ON c.subject_code_id=s.subject_code_id
                WHERE s.student_id=$student_id AND s.week_id = $current_week_id
                $sql_inject    
                GROUP BY s.subject_code_id ";

但是，@total_ Correct、@total_target 的值在 php mysql 执行时返回 null 。 当我在mysql IDE中运行时，结果是好的。

如何解决这个问题？

Answer 1

这是正确的原因，要使其工作，您需要使用SELECT INTO... 构造，例如select col1 into @arg1 from tbl1。此外，既然您是从 PHP 运行查询，为什么还需要它呢？如果您确实需要，请考虑将查询包装在存储过程中，并将这些参数作为 OUT 参数。

嗯，它返回 null，因为您没有选择这些参数。查询执行后，您需要选择这些参数:select @total_ Correct, @total_target;。

为什么不按原样运行查询(如下所示)并获取特定列的值

SELECT 
                c.subject_code_id, c.subject_name, SUM(s.total_target_question) AS total_target_question,
                ROUND((RAND() * (@total_target-10))+10) AS total_correct, 
                (@total_correct / @total_target)*100 AS percent, c.icon_filename
                 FROM 
                edu_subject_code c LEFT JOIN wkp_wg_student_subject s ON c.subject_code_id=s.subject_code_id
                WHERE s.student_id=$student_id AND s.week_id = $current_week_id
                $sql_inject    
                GROUP BY s.subject_code_id