这会拉回昨天和今天的两个int值。我想从第三列中的语句中减去两个结果，称为差异：
SELECT (
SELECT COUNT(*)
FROM collectors_users
WHERE DATE(dateadded) = CURDATE() - INTERVAL 1 DAY
) AS yesterday, COUNT(*) AS today
FROM collectors_users
WHERE DATE(dateadded) = CURDATE()
您需要重复这些表达式。SQL（通常）不允许您在same中重复使用列别名SELECT。您可以将逻辑简化为：
SELECT SUM(DATE(dateadded) = CURDATE() - INTERVAL 1 DAY) AS yesterday,
SUM(DATE(dateadded) = CURDATE()) as today,
(SUM(DATE(dateadded) = CURDATE()) -
SUM(DATE(dateadded) = CURDATE() - INTERVAL 1 DAY)
) as diff
FROM collectors_users
WHERE dateadded >= CURDATE() - INTERVAL 1 DAY AND
dateadded < CURDATE() + INTERVAL 1 DAY;
请注意，该WHERE子句的逻辑为期两天。另外，它不使用DATE()。这将允许查询使用索引（如果有）。
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