回答了，为什么呢？ Mysql的第一个查询运行第二个不需要理解为什么？

Question

此SQL查询有效，但第二个查询不能有人解释为什么和/或使其工作吗？

第一个基于行选择，第二个我希望所有列和行更多地具有全局平均值。该名称可能有1到X个条目，因此必须按名称分组。

SELECT NAME,
       AVERAGE,
       AD1, AD2, AD3, AD4, AD5, AD6,
       (AD1 + AD2 + AD3 + AD4 + AD5 + AD6) AS TOTAL_AD
FROM ( SELECT
    NAME,
    AVERAGE,
    (AVG1-AVERAGE) AS AD1,
    (AVG2-AVERAGE) AS AD2,
    (AVG3-AVERAGE) AS AD3,
    (AVG4-AVERAGE) AS AD4,
    (AVG5-AVERAGE) AS AD5,
    (AVG6-AVERAGE) AS AD6
    FROM P_AVG ) AS T
ORDER BY NAME;

按预期运行。

SELECT NAME,
       AVG(AVERAGE),
       AD1, AD2, AD3, AD4, AD5, AD6,
       (AD1 + AD2 + AD3 + AD4 + AD5 + AD6) AS TOTAL_AD
FROM ( SELECT
    NAME,
    AVG(AVERAGE),
    (AVG(AVG1)-AVG(AVERAGE)) AS AD1,
    (AVG(AVG2)-AVG(AVERAGE)) AS AD2,
    (AVG(AVG3)-AVG(AVERAGE)) AS AD3,
    (AVG(AVG4)-AVG(AVERAGE)) AS AD4,
    (AVG(AVG5)-AVG(AVERAGE)) AS AD5,
    (AVG(AVG6)-AVG(AVERAGE)) AS AD6
    FROM P_AVG GROUP BY NAME ) AS T;

对于该列所在的查询，MySQL返回此错误。

1054-“字段列表”中的未知列“ AVERAGE”

不运行并返回错误。

此问题已解决，但为什么呢？

SELECT NAME,
       AD,
       AD1, AD2, AD3, AD4, AD5, AD6,
       (AD1 + AD2 + AD3 + AD4 + AD5 + AD6) AS TOTAL_AD
FROM ( SELECT
    NAME,
    (AVG(AVERAGE)) AS AD,
    (AVG(AVG1)-AVG(AVERAGE)) AS AD1,
    (AVG(AVG2)-AVG(AVERAGE)) AS AD2,
    (AVG(AVG3)-AVG(AVERAGE)) AS AD3,
    (AVG(AVG4)-AVG(AVERAGE)) AS AD4,
    (AVG(AVG5)-AVG(AVERAGE)) AS AD5,
    (AVG(AVG6)-AVG(AVERAGE)) AS AD6
    FROM P_AVG GROUP BY NAME ) AS T;

Answer 1

查询中的错误是由于缺少(AVG(AVERAGE))

的适当别名引起的

通常，您是通过外部选择不存在的列名

尝试在此列中添加适当的别名广告，例如：

  AVG(AVERAGE) AVERAGE

。

  SELECT NAME,
         AVG(AVERAGE),
         AD1, AD2, AD3, AD4, AD5, AD6,
         (AD1 + AD2 + AD3 + AD4 + AD5 + AD6) AS TOTAL_AD
  FROM ( 
      SELECT
      NAME,
      AVG(AVERAGE) AVERAGE,
      (AVG(AVG1)-AVG(AVERAGE)) AS AD1,
      (AVG(AVG2)-AVG(AVERAGE)) AS AD2,
      (AVG(AVG3)-AVG(AVERAGE)) AS AD3,
      (AVG(AVG4)-AVG(AVERAGE)) AS AD4,
      (AVG(AVG5)-AVG(AVERAGE)) AS AD5,
      (AVG(AVG6)-AVG(AVERAGE)) AS AD6
      FROM P_AVG 
      GROUP BY NAME ) AS T;