有什么方法可以使用“或” Google Directory API users.list查询中的操作符? [英] Is there any way to use an "or" operator in Google Directory API users.list query?
问题描述
我想使用 users.list
我写了查询:
givenName:m *或familyName:m *
,但它没有工作。
只有 givenName:m * familyName:m *
可以工作,但它在子句中插入和运算符。
In docs逗留:
$ b
多个子句由空格分隔,并通过和运算符隐式地加入。
我只能用隐式和运算符编写查询吗?有什么办法如何使用或?来编写它? 解决方案
API 我发现我真的不'您需要 {{PREFIX}} * 或一些运营商来完成我的目标。
如果您想搜索所有给定或姓氏开始的用户,例如在字母m上,只需在查询参数中放入一个 m
(没有字段名称,例如givenName:)。看来,API隐式搜索所有域用户的 givenName 和 familyName 字段。它还搜索所有电子邮件字段的子字符串(例如 martin @ example.com)中的子字符串(@之前)。
在查询中使用单词example,它会为所有用户提供电子邮件字段,如@ example .com。
如果您只放入了一些没有指定字段的字符串,那么API也可能会在其他字段中进行搜索,但我没有找到。
我不知道如果这是一个错误或一个功能,但我知道它部分回答了我的问题。
I want to build a search query for Google domain users using users.list Google Directory API. I tried to build a query according docs, which will select all users with givenName or familyName starts e.g. on letter "m".
I wrote the query:
givenName:m* or familyName:m*
, but it doesn't work.
Only givenName:m* familyName:m*
works but it inserts "and" operator between clauses.
In docs stays:
Multiple clauses are separated by whitespace and are implicitly joined by an "and" operator.
Can I only write a query with the implicit "and" operator? Is there any way how to write it using "or"?
After some time experimenting with the API I have found out that I really don't need {{PREFIX}}* or some operators to accomplish my goal.
If you want to search all the users whose given or family name starts e.g. on letter "m", just put a single m
(without field name e.g. "givenName:") into the query parameter. It seems that API implicitly searches in all domain users' givenName and familyName fields. It also searches in a substring (before "@") of all email fields (e.g. martin@example.com).
If you put word "example" in the query, it'll gives you all users with email field something@example.com.
It is possible that if you put only some string without any specified fields, the API may also searches in any other field, but I didn't find in which.
I don't know if this is a bug or a feature, but I know it partially gives an answer to my question.
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