Question

我希望获得与sql签订合同的唯一，at_least_one和总计数。下面的代码抛出一个错误： 标量子查询产生了多个元素

#standardSQL
WITH estrato_contrato AS (SELECT ESTRATO, CONTRATO, count(*) AS count FROM `mytable` GROUP BY ESTRATO, CONTRATO)


SELECT ESTRATO, COUNT(DISTINCT CONTRATO) AS only_once_contract, sum(count) AS total,
(SELECT COUNT(DISTINCT CONTRATO) FROM estrato_contrato WHERE count > 1 GROUP BY ESTRATO) AS at_least_one
FROM estrato_contrato
GROUP BY ESTRATO

Answer 1

制定子查询的正确方法是使用关联子句：

SELECT ESTRATO, COUNT(DISTINCT CONTRATO) AS only_once_contract, sum(count) AS total,
       (SELECT COUNT(DISTINCT ec2.CONTRATO)
        FROM estrato_contrato ec2
        WHERE ec2.count > 1
              ec2.ESTRATO = ec.ESTRATO
       ) AS at_least_one
FROM estrato_contrato ec
GROUP BY ESTRATO;

可能还有其他方法来编写查询，但这是您的版本的意图。

Answer 2

试试这个

WITH estrato_contrato AS（SELECT ESTRATO，CONTRATO，count（*）AS count FROM EStable GROUP BY ESTRATO，CONTRATO）

SELECT ESTRATO，COUNT（DISTINCT CONTRATO）AS only_once_contract，sum（count）AS total，CASE WHEN COUNT（DISTINCT CONTRATO）＆gt; 1 then 1 else 0 end as as_least_one 来自estrato_contrato GROUP BY ESTRATO

Answer 3

以下是BigQuery Standard SQL

没有子选择的选项

#standardSQL
WITH estrato_contrato AS (
  SELECT ESTRATO, CONTRATO, COUNT(*) AS COUNT FROM  `mytable`  GROUP BY ESTRATO, CONTRATO
)
SELECT 
  ESTRATO, 
  COUNT(CONTRATO) AS only_once_contract, 
  SUM(COUNT) AS total,
  ARRAY_LENGTH(ARRAY_AGG(IF(COUNT > 1, CONTRATO, NULL) IGNORE NULLS)) AS at_least_one
FROM estrato_contrato ec
GROUP BY ESTRATO

最后，最简单的是（注意：你不必担心外部选择中的不同值，因为它已经在estrato_contrato中处理了

#standardSQL
WITH estrato_contrato AS (
  SELECT ESTRATO, CONTRATO, COUNT(*) AS COUNT FROM  `mytable` GROUP BY ESTRATO, CONTRATO
)
SELECT 
  ESTRATO, 
  COUNT(CONTRATO) AS only_once_contract, 
  SUM(COUNT) AS total,
  COUNTIF(COUNT > 1) AS at_least_one
FROM estrato_contrato ec
GROUP BY ESTRATO

子查询产生了多个元素

3 个答案: