Question

SELECT January.customer_id AS Jancust_id, 
COUNT( DISTINCT January.customer_id ) AS Jan_orig_cust,
COUNT (DISTINCT payments.customer_id) as Jan_retain_cust,
Jan_retain_cust/ 1533 AS CAC_referral
FROM January_Cohort January
LEFT JOIN telemon_payments_data payments
ON January.customer_id = payments.customer_id
WHERE January.acquisition_source = 'referral'
AND 
payments.payment_date BETWEEN '2016-06-01' and '2016-06-31'

好的，我不能对使用COUNT＆amp;的同一查询中的列进行数学运算。刚刚结束？错误信息如下。

“＃1064 - 您的SQL语法出错;请查看手册对应于您的MySQL服务器版本，以便使用正确的语法靠近'DISTINCT payments.customer_id）的Jan_retain_cust， Jan_retain_cust / 1533 AS CAC_'在第3行“

Answer 1

您缺少group by条款。此外，您尝试在 not 允许的同一选择中的表达式中使用别名Jan_retain_cust。只需使用原始表达式Jan_retain_cust替换别名COUNT (DISTINCT payments.customer_id)。

select January.customer_id as Jancust_id,
    COUNT(distinct January.customer_id) as Jan_orig_cust,
    COUNT(distinct payments.customer_id) as Jan_retain_cust,
    COUNT(distinct payments.customer_id) / 1533 as CAC_referral
from January_Cohort January
left join telemon_payments_data payments on January.customer_id = payments.customer_id
where January.acquisition_source = 'referral'
    and payments.payment_date between '2016-06-01' and '2016-06-31'
group by January.customer_id;

Answer 2

您不能在同一查询中引用别名列。

SELECT January.customer_id AS Jancust_id, 
COUNT( DISTINCT January.customer_id ) AS Jan_orig_cust,
COUNT (DISTINCT payments.customer_id) as Jan_retain_cust,
Jan_retain_cust/ 1533 AS CAC_referral    <--- You can not do.
FROM January_Cohort January
LEFT JOIN telemon_payments_data payments
ON January.customer_id = payments.customer_id
WHERE January.acquisition_source = 'referral'
AND 
payments.payment_date BETWEEN '2016-06-01' and '2016-06-31'

另外，正如@GurV指出的那样，你也缺少group by条款。

SQL：在COUNT列上进行数学运算

2 个答案: