Question

我有以下查询，它向我提供了项目的整体状态。我想知道是否有更好的方法来编写相同的查询。

表的示例架构：

CREATE TABLE `test_status`
(
      `test_id` int
(11) NOT NULL,
      `group_id` int
(11) NOT NULL,
      `sub_test_id` int
(11) NOT NULL,
      `project_id` int
(11) NOT NULL,
      `collection_status` varchar
(20) DEFAULT NULL,
      `labeling_status` varchar
(20) DEFAULT NULL,
      `upload_status` varchar
(20) DEFAULT NULL,
      `upload_date` date DEFAULT NULL,
      `disk_number` int
(11) DEFAULT NULL,
      `subject_id` varchar
(10) DEFAULT NULL,
      `collection_id` varchar
(25) DEFAULT NULL,
      `assigned_to` int
(11) DEFAULT NULL,
      `assigned_on` date DEFAULT NULL,
      `turned_in_date` date DEFAULT NULL,
      PRIMARY KEY
(`test_id`,`group_id`,`sub_test_id`,`project_id`));

我的查询

select
    (select count(*)
    from test_status
    where project_id = 7 and collection_status = 'COLLECTED') as collected,
    (select count(*)
    from test_status
    where project_id = 7 and collection_status = 'SCHEDULED') as scheduled,
    (select count(*)
    from test_status
    where project_id = 7 and collection_status = 'CORRUPTED') as corrupted,
    (select count(*)
    from test_status
    where project_id = 7 and collection_status is NULL) as 'not collected',
    (select count(*)
    from test_status
    where project_id = 7 and ((labeling_status = 'GOOD' and (upload_status != 'UPLOADED' or upload_status != 'QUEUED')) or (labeling_status = 'Labeled'))) as 'Labeled',
    (select count(*)
    from test_status
    where project_id = 7 and (labeling_status = 'RAW') or (labeling_status = 'ASSIGNED') ) as 'to be labeled',
    (select count(*)
    from test_status
    where project_id = 7 and labeling_status = 'RE-LABEL') as 'Re-label',
    (select count(*)
    from test_status
    where project_id = 7 and (upload_status = 'UPLOADED' or upload_status = 'QUEUED')) as 'Uploaded',
    (select count(*)
    from test_status
    where project_id = 7 and labeling_status = 'GOOD' and upload_status is null) as 'ready to be uploaded';

Answer 1

您可以尝试使用条件汇总函数会更好。

在CASE WHEN中设置条件

SELECT 
    COUNT(CASE WHEN collection_status = 'COLLECTED' THEN 1 END),
    COUNT(CASE WHEN collection_status = 'SCHEDULED' THEN 1 END),
    COUNT(CASE WHEN collection_status = 'CORRUPTED' THEN 1 END),
    COUNT(CASE WHEN collection_status is NULL THEN 1 END),
    COUNT(CASE WHEN  ((labeling_status = 'GOOD' and (upload_status != 'UPLOADED' or upload_status != 'QUEUED')) or (labeling_status = 'Labeled'))THEN 1 END),
    COUNT(CASE WHEN (labeling_status = 'RAW') or (labeling_status = 'ASSIGNED') THEN 1 END),
    COUNT(CASE WHEN labeling_status = 'RE-LABEL' THEN 1 END),
    COUNT(CASE WHEN (upload_status = 'UPLOADED' or upload_status = 'QUEUED') THEN 1 END),
    COUNT(CASE WHEN labeling_status = 'GOOD' and upload_status is null THEN 1 END)
FROM test_status
WHERE project_id = 7

或者您可以尝试一种更简单的方法，将SUM（0或1）与bool的{{1}}

count

有没有更好的方法来编写我编写的查询

1 个答案: