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我正在将此代码写入一个PHP变量,其中$ username和$ email被替换。我将它回显到屏幕上,它看起来很正确。我假设MySQL代码存在问题,因为我总是得到的唯一结果是“可接受的”。任何帮助?找不到为什么每个查询都返回else(case when)
SELECT CASE
WHEN email='$email' THEN '$email is already associated with an account'
WHEN username='$username' THEN '$username is already taken'
ELSE 'acceptable'
END AS result FROM collaborator
UNION
SELECT CASE
WHEN email='$email' THEN '$email is already associated with an account'
WHEN username='$username' THEN '$username is already taken'
ELSE 'acceptable'
END AS result
FROM waitForValidation
LIMIT 1;
可能是区分大小写的问题?在所有情况下,工会是否会返回“不可接受”的结果?因此返回的顺序可能是可变返回限制1,从而消除了正确的值? – xQbert