我会回答，但为什么？Mysql第一次查询运行第二次不需要理解为什么？

这个SQL查询有效，但第二个查询无效。有人能解释一下为什么和/或如何使它有效吗

第一个是基于行的选择，第二个是我希望所有列和行更接近全局平均值。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;

Mysql为列所在的查询返回此错误

1054-未知列“字段列表”中的“平均值” 不运行并返回错误

这解决了问题，但为什么

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;

---

在您的查询中，错误是由于缺少AVGAVERAGE的正确别名造成的

实际上，您正在通过外部选择一个不存在的列名

尝试在此列中添加适当的别名，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;