1

这是我在 MySQL 中使用的示例。但是，在 BigQuery 中，我的 OnSitetimestamp是DATE而我的 Documents 时间戳是TIMESTAMP。

BigQuery 在以下查询中遇到问题，因为我收到以下消息：

参数类型的函数 DATE 没有匹配的签名：DATE。支持的签名：DATE(TIMESTAMP, [STRING])；日期（日期时间）；日期（INT64，INT64，INT64）在 [8:146]

有谁知道我需要做什么才能使查询与比较 DATE 和 TIMESTAMP 一起工作？

架构（MySQL v5.7）

CREATE TABLE OnSite
    (`uid` varchar(55), `worksite_id`  varchar(55), `timestamp` datetime)
;

INSERT INTO OnSite
    (`uid`, `worksite_id`, `timestamp`)
VALUES
  ("u12345", "worksite_1", '2019-01-01'),
  ("u12345", "worksite_1", '2019-01-02'),
  ("u12345", "worksite_1", '2019-01-03'),
  ("u12345", "worksite_1", '2019-01-04'),
  ("u12345", "worksite_1", '2019-01-05'),
  ("u12345", "worksite_1", '2019-01-06'),
  ("u1", "worksite_1", '2019-01-01'),
  ("u1", "worksite_1", '2019-01-02'),
  ("u1", "worksite_1", '2019-01-05'),
  ("u1", "worksite_1", '2019-01-06')

;


CREATE TABLE Documents
    (`document_id` varchar(55), `uid` varchar(55), `worksite_id`  varchar(55), `type` varchar(55), `timestamp` datetime)
;

INSERT INTO Documents
    (`document_id`, `uid`, `worksite_id`, `type`, `timestamp`)

VALUES
  ("1",     "u12345",   "worksite_1", 'work_permit',    '2019-01-01 00:00:00'),
  ("2",     "u12345",   "worksite_2", 'job',            '2019-01-02 00:00:00'),
  ("3",     "u12345",   "worksite_1", 'work_permit',    '2019-01-03 00:00:00'),
  ("4",     "u12345",   "worksite_2", 'job',            '2019-01-04 00:00:00'),
  ("5",     "u12345",   "worksite_1", 'work_permit',    '2019-01-05 00:00:00'),
  ("6",     "u12345",   "worksite_2", 'job',            '2019-01-06 00:00:00'),
  ("7",     "u12345",   "worksite_1", 'work_permit',    '2019-01-07 00:00:00'),
  ("8",     "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00'),
  ("9",     "u12345",   "worksite_1", 'job',            '2019-01-09 00:00:00'),
  ("10",    "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00'),
  ("11",    "u12345",   "worksite_1", 'work_permit',    '2019-01-09 00:00:00'),
  ("12",    "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00'),
  ("13",    "u12345",   "worksite_1", 'job',            '2019-01-09 00:00:00'),
  ("14",    "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00'),
  ("15",    "u12345",   "worksite_1", 'work_permit',    '2019-01-09 00:00:00')

;

查询 #1

SELECT
  IFNULL(OnSite.worksite_id, Documents.worksite_id) as `Worksite`,
  DATE(IFNULL(OnSite.timestamp, Documents.timestamp)) as `Date`,
  COUNT(Documents.worksite_id) as `Users_on_Site`,
  COUNT(DISTINCT OnSite.uid) as `Completed`

FROM OnSite
  LEFT JOIN Documents ON OnSite.worksite_id = Documents.worksite_id AND DATE(OnSite.timestamp) = DATE(Documents.timestamp)
GROUP BY `Date`, `Worksite`;

| Worksite   | Date       | Users_on_Site | Completed |
| ---------- | ---------- | ------------- | --------- |
| worksite_1 | 2019-01-01 | 2             | 2         |
| worksite_1 | 2019-01-02 | 0             | 2         |
| worksite_1 | 2019-01-03 | 1             | 1         |
| worksite_1 | 2019-01-04 | 0             | 1         |
| worksite_1 | 2019-01-05 | 2             | 2         |
| worksite_1 | 2019-01-06 | 0             | 2         |

在 DB Fiddle 上查看

4

3 回答 3

2

以下是 BigQuery 标准 SQL 

#standardSQL
SELECT
  IFNULL(OnSite.worksite_id, Documents.worksite_id) AS `Worksite`,
  IFNULL(OnSite.timestamp, DATE(Documents.timestamp)) AS `DATE`,
  COUNT(Documents.worksite_id) AS `Users_on_Site`,
  COUNT(DISTINCT OnSite.uid) AS `Completed`
FROM `project.dataset.OnSite` OnSite
LEFT JOIN `project.dataset.Documents` Documents 
ON OnSite.worksite_id = Documents.worksite_id 
AND OnSite.timestamp = DATE(Documents.timestamp)
GROUP BY `DATE`, `Worksite`

是否适用于您问题的样本数据 

WITH `project.dataset.OnSite` AS (
  SELECT "u12345" uid, "worksite_1" worksite_id, DATE '2019-01-01' `TIMESTAMP` UNION ALL
  SELECT "u12345", "worksite_1", '2019-01-02' UNION ALL
  SELECT "u12345", "worksite_1", '2019-01-03' UNION ALL
  SELECT "u12345", "worksite_1", '2019-01-04' UNION ALL
  SELECT "u12345", "worksite_1", '2019-01-05' UNION ALL
  SELECT "u12345", "worksite_1", '2019-01-06' UNION ALL
  SELECT "u1", "worksite_1", '2019-01-01' UNION ALL
  SELECT "u1", "worksite_1", '2019-01-02' UNION ALL
  SELECT "u1", "worksite_1", '2019-01-05' UNION ALL
  SELECT "u1", "worksite_1", '2019-01-06' 
), `project.dataset.Documents` AS (
  SELECT "1" document_id,     "u12345" uid,   "worksite_1" worksite_id, 'work_permit' type,    TIMESTAMP '2019-01-01 00:00:00' `TIMESTAMP` UNION ALL
  SELECT "2",     "u12345",   "worksite_2", 'job',            '2019-01-02 00:00:00' UNION ALL
  SELECT "3",     "u12345",   "worksite_1", 'work_permit',    '2019-01-03 00:00:00' UNION ALL
  SELECT "4",     "u12345",   "worksite_2", 'job',            '2019-01-04 00:00:00' UNION ALL
  SELECT "5",     "u12345",   "worksite_1", 'work_permit',    '2019-01-05 00:00:00' UNION ALL
  SELECT "6",     "u12345",   "worksite_2", 'job',            '2019-01-06 00:00:00' UNION ALL
  SELECT "7",     "u12345",   "worksite_1", 'work_permit',    '2019-01-07 00:00:00' UNION ALL
  SELECT "8",     "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00' UNION ALL
  SELECT "9",     "u12345",   "worksite_1", 'job',            '2019-01-09 00:00:00' UNION ALL
  SELECT "10",    "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00' UNION ALL
  SELECT "11",    "u12345",   "worksite_1", 'work_permit',    '2019-01-09 00:00:00' UNION ALL
  SELECT "12",    "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00' UNION ALL
  SELECT "13",    "u12345",   "worksite_1", 'job',            '2019-01-09 00:00:00' UNION ALL
  SELECT "14",    "u12345",   "worksite_2", 'work_permit',    '2019-01-09 00:00:00' UNION ALL
  SELECT "15",    "u12345",   "worksite_1", 'work_permit',    '2019-01-09 00:00:00' 
)

结果将如预期 

Row Worksite    Date        Users_on_Site   Completed    
1   worksite_1  2019-01-01  2               2    
2   worksite_1  2019-01-02  0               2    
3   worksite_1  2019-01-03  1               1    
4   worksite_1  2019-01-04  0               1    
5   worksite_1  2019-01-05  2               2    
6   worksite_1  2019-01-06  0               2
于 2019-01-17T20:07:24.233 回答
1

在BigQuery 文档中，解释了该DATE函数接受以下输入：

  1. DATE(year, month, day): 从表示年、月和日的 INT64 值构造 DATE。

  2. DATE(timestamp_expression[, timezone])：将 timestamp_expression 转换为 DATE 数据类型。它支持一个可选参数来指定一个时区。如果未指定时区，则使用默认时区 UTC。

在您的用例中，您传递给的值似乎DATE已经是日期时间。为此，您可以使用DATETIME_TRUNC，例如：

DATETIME_TRUNC(IFNULL(OnSite.timestamp, Documents.timestamp), DAY)
于 2019-01-17T15:29:31.590 回答
0

你为什么不强迫一切，让生活更轻松:-)？所有这些都应该起作用：

select 
   date(timestamp('2019-01-02')), 
   date(timestamp('2019-01-02 00:00:00')), 
   date(timestamp(null)))

因此，在您的 if null 语句中：

SELECT
  IFNULL(OnSite.worksite_id, Documents.worksite_id) as `Worksite`,
  IFNULL(date(datetime(OnSite.timestamp)),date(datetime(Documents.timestamp))) as `Date`,
  COUNT(Documents.worksite_id) as `Users_on_Site`,
  COUNT(DISTINCT OnSite.uid) as `Completed`
FROM OnSite
  LEFT JOIN Documents ON OnSite.worksite_id = Documents.worksite_id AND DATE(datetime(OnSite.timestamp)) = DATE(datetime(Documents.timestamp))
GROUP BY `Date`, `Worksite`;
于 2019-01-17T20:16:40.153 回答