我试图获取数据库中特定文件夹中的项目总数 - SQL(HP ALM)

时间:2016-01-08 22:09:43

标签: mysql sql alm qc

我试图在HP ALM 11的“测试计划”模块中获取每个应用程序文件夹中的测试用例总数。

这是我到目前为止所做的:

SELECT
Count (TS_TEST_ID) As "App1",
Count (TS_TEST_ID) As "App2",
Count (TS_TEST_ID) As "App3",
Count (TS_TEST_ID) As "App4",
Count (TS_TEST_ID) As "App5",
Count (TS_TEST_ID) As "App6",
Count (TS_TEST_ID) As "App7",
Count (TS_TEST_ID) As "App8"

FROM DESSTEPS, TEST, ALL_LISTS
WHERE DS_TEST_ID = TS_TEST_ID
AND  AL_ITEM_ID = TS_SUBJECT
AND TS_SUBJECT = ALL_LISTS.AL_ITEM_ID
AND AL_DESCRIPTION = 'App1 Folder Name in the Database'

查询为每个应用程序名称输出一列,但我只能得到App1的测试用例总数。让我们说这个数字是100.它为每个应用程序全面挑出100个。有没有办法列出所有应用程序文件夹,以便每列有与该应用程序相关的总计数?我需要以某种方式列出所有文件夹名称,我无法弄明白。这可能吗?感谢您的帮助。

1 个答案:

答案 0 :(得分:1)

所以,你想要每个应用程序的测试数量,每个AL_DESCRIPTION代表一个应用程序,你想要8个特定的应用程序,对吗?我想你想要的是以下内容:

SELECT AL_DESCRIPTION, Count (TS_TEST_ID)
FROM DESSTEPS, TEST, ALL_LISTS
WHERE DS_TEST_ID = TS_TEST_ID
  AND  AL_ITEM_ID = TS_SUBJECT
  AND TS_SUBJECT = ALL_LISTS.AL_ITEM_ID
  AND AL_DESCRIPTION IN
    ('App1 Folder Name in the Database', 'App2 Folder Name in the Database', [...],'App8 Folder Name in the Database')
GROUP BY AL_DESCRIPTION

但值会垂直显示,但我不认为这是一个问题,是吗?