Question

我想用变量替换字段名来从PHP和MySQL中的表中获取数据。

select * from demo where variable-1

而不是select * from demo where fieldname=1

因为从drowpdown中选择了超过50个fieldname。

Answer 1

在field names以及table Names上

变量不起作用。它唯一能做的就是创建动态sql ，例如

SET @variableName = 'fieldname';
SET @sql = CONCAT('SELECT * FROM demo WHERE `', @variableName, '` = 1');
PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

Dynamic SQL (Prepared Statements)

Answer 2

您可以使用Prepared Statements

delimiter // 
CREATE PROCEDURE dynamic(IN tbl CHAR(64), IN col CHAR(64))
BEGIN
    SET @s = CONCAT('SELECT ',col,' FROM ',tbl );
    PREPARE stmt FROM @s;
    EXECUTE stmt;
END
//
delimiter ;

在选择查询中用变量替换字段名称

2 个答案: