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MySQL - 在第二个动态列中引用动态列

转载 作者:行者123 更新时间:2023-11-29 07:11:20 26 4
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我有一个这样的表:

user | test_score
a | 50
b | 60

我需要查询结果,其中有一列是测试分数的 1.5 倍,然后是与原始分数的差异。结果将是:

user | test_score | adjusted_test_score | test_score_diff
a | 50 | 75 | 25
b | 60 | 90 | 30

我试过的查询:

 SELECT
user AS user,
test_score AS test_score,
(test_score * 1.5) AS adjusted_test_score,
(adjusted_test_score - test_score) AS test_score_diff

返回“adjusted_test_score”列不存在的错误。

有没有办法在不进行连接的情况下引用此列?

最佳答案

尝试:

SELECT user, test_score, adjusted_test_score,
(adjusted_test_score - test_score) AS test_score_diff
FROM (
SELECT user, test_score, (test_score * 1.5) AS adjusted_test_score
FROM source_table
)

或者你也可以这样做:

SELECT user, test_score, (test_score * 1.5) AS adjusted_test_score,
((test_score * 1.5) - test_score) AS test_score_diff
FROM source_table

关于MySQL - 在第二个动态列中引用动态列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4287732/

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