SELECT (DISTINCT visits.user_id), purchases FROM visits
尝试这样做,但它不起作用。
答案 0 :(得分:-1)
使用聚合。也许:
SELECT v.user_id, ANY_VALUE(purchases)
FROM visits v
GROUP BY v.user_id;
或者,如果您想要购买的总数——假设 purchases
是一个数字:
SELECT v.user_id, SUM(purchases)
FROM visits v
GROUP BY v.user_id;