调用Graph API /me/calendar/getSchedule获取会议室日程返回空数组问题
问题:调用Microsoft Graph API的/me/calendar/getSchedule获取会议室日程为空
我在Graph Explorer中调用Microsoft Graph API的/me/calendar/getSchedule端点,获取指定会议室room@mail.com的日程。明明该时间段内存在包含此会议室的事件,但返回的scheduleInformation中scheduleItems数组为空。测试查询创建该事件的用户日程时,结果正常。请问可能的原因是什么?
请求体
{ "schedules": [ "room@mail.com" ], "startTime": { "dateTime": "2022-12-11T09:00:00", "timeZone": "UTC" }, "endTime": { "dateTime": "2022-12-13T23:59:00", "timeZone": "UTC" } }
响应体
{ "@odata.context": "https://graph.microsoft.com/v1.0/$metadata#Collection(microsoft.graph.scheduleInformation)", "value": [ { "scheduleId": "room@mail.com", "availabilityView": "000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000", "scheduleItems": [] } ] }
创建的事件示例
{ "@odata.etag": "W/\"1z8A//yOl0a9+ObgH9y4KQAAAfmk/Q==\"", "id": "some id", "subject": "Let's go for lunch", "bodyPreview": "Does mid month work for you?", "body": { "contentType": "html", "content": "<html><head><meta http-equiv=\"Content-Type\" content=\"text/html; charset=utf-8\"></head><body>Does mid month work for you?</body></html>" }, "start": { "dateTime": "2022-12-13T12:00:00.0000000", "timeZone": "UTC" }, "end": { "dateTime": "2022-12-14T14:00:00.0000000", "timeZone": "UTC" }, "location": { "displayName": "room@mail.com", "locationType": "default", "uniqueId": "room@mail.com", "uniqueIdType": "private" }, "attendees": [], "organizer": { "emailAddress": { "name": "admin", "address": "admin@mail.com" } } }
可能的原因
- 事件未将会议室作为参会者添加:从提供的事件示例可见,
attendees数组为空,仅在location字段填写了会议室邮箱。getSchedule接口需要会议室被添加为参会者(优先是必需参会者),才能识别其日程占用状态,仅设置位置不会被统计到会议室的日程中。 - 权限不足:当前调用接口的账号没有读取该会议室日历的权限。需确保账号已授权
Calendars.Read或Calendars.Read.Shared权限,且该会议室日历已共享给该账号,或者账号具备全局管理员/日历管理员权限。 - 会议室邮箱配置异常:该会议室邮箱可能未被正确配置为Exchange资源邮箱,或者其日历权限设置中限制了查询账号的访问。需在Exchange管理中心检查该会议室的日历共享权限,确保允许查询账号读取其日程。
- 时间范围覆盖不全:请求的结束时间是
2022-12-13T23:59:00(UTC),而事件结束时间是2022-12-14T14:00:00(UTC),不过事件开始时间在请求范围内,这一点大概率不是主因,但仍需确认时间转换是否存在误差。
内容的提问来源于stack exchange,提问作者Ewelin
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