mysql - 未获取以变量作为 json 键的 Mysql json 值

Question

SET @im := (SELECT items->"$.one" From fruits where id = 12);
//it's working and saved result into @im variable

但我需要用函数中的变量来获取它 我尝试过但失败了.. 我尝试了 concat，它也不起作用， 我尝试准备执行 stmt，它也不起作用。

SET @p := "one";
SET @im := (SELECT items->"$."@p From fruits where id = 12);
//it's not working


"$.one" ==> "$."@p
//difference..

帮帮我，有什么办法可以得到这个..

Answer 1

尝试一下这个方法，应该可行。

SET @p = 'one';
SET @sql = '(SELECT items->"$.@p" into @im From fruits where id = 12)';
SET @sql = replace(@sql,'@p',@p);
PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
select @im;