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To add an user to the existing group

3004
Member Posts: 204,171 Green Ribbon
The following two xml work fine when i create
a user and add it to a group.
But when i try to add a new user to the existing group, it creates another group
with the same name.How come iFS accepts
groups with the same name?
Can anyone spare your time to answer this question.
<?xml version = '1.0' standalone = 'yes'?>
<SimpleUser>
<UserName>msmith</UserName>
<Password>ifs</Password>
<DistinguishedNameSuffix>.ambiguity.com</DistinguishedNameSuffix>
<AdminEnabled>false</AdminEnabled>
<HomeFolderRoot>/home</HomeFolderRoot>
<EmailAddressSuffix>@ambiguity.com</EmailAddressSuffix>
</SimpleUser>
<?xml version = '1.0' standalone = 'yes'?>
<DIRECTORYGROUP>
<Name>DemonstrationUsers</Name>
<Members>
<REF reftype='name'>msmith</REF>
</Members>
</DIRECTORYGROUP>
Thanks for your answer in advance
a user and add it to a group.
But when i try to add a new user to the existing group, it creates another group
with the same name.How come iFS accepts
groups with the same name?
Can anyone spare your time to answer this question.
<?xml version = '1.0' standalone = 'yes'?>
<SimpleUser>
<UserName>msmith</UserName>
<Password>ifs</Password>
<DistinguishedNameSuffix>.ambiguity.com</DistinguishedNameSuffix>
<AdminEnabled>false</AdminEnabled>
<HomeFolderRoot>/home</HomeFolderRoot>
<EmailAddressSuffix>@ambiguity.com</EmailAddressSuffix>
</SimpleUser>
<?xml version = '1.0' standalone = 'yes'?>
<DIRECTORYGROUP>
<Name>DemonstrationUsers</Name>
<Members>
<REF reftype='name'>msmith</REF>
</Members>
</DIRECTORYGROUP>
Thanks for your answer in advance
Comments
-
Working from home so I cannot verify
Try
<?xml version = '1.0' standalone = 'yes'?>
<DIRECTORYGROUP>
<Update reftype="Name">DemonstrationUsers</Update>
<Members>
<REF reftype='name'>msmith</REF>
</Members>
</DIRECTORYGROUP>
null -
Thanks again.
It really worked. But how to find this <Update> tags and all other stuff.
We are totally at dark and had to post our queries only in the forum.
This discussion has been closed.