Hi All |
Re: Name of the source module without read rights You cannot get a handle on incoming 'Link's without opening the source module, which you cannot do without Read rights. Perhaps this will work:
ModName_ mn
for mn in obj <-"*" do
{ print fullName(mn) "\n"
}
or perhaps this:
LinkRef lr
for lr in obj <-"*" do
{ string Name1 = source(lr)
ModName_ mn = source(lr)
string Name2 = name(mn)
print Name1 "\t" Name2 "\n"
}
|
Re: Name of the source module without read rights llandale - Wed Aug 12 14:05:36 EDT 2009 You cannot get a handle on incoming 'Link's without opening the source module, which you cannot do without Read rights. Perhaps this will work:
ModName_ mn
for mn in obj <-"*" do
{ print fullName(mn) "\n"
}
or perhaps this:
LinkRef lr
for lr in obj <-"*" do
{ string Name1 = source(lr)
ModName_ mn = source(lr)
string Name2 = name(mn)
print Name1 "\t" Name2 "\n"
}
I agree with that but I just wanted the name of the source module using linkref. I didn't want any handle Regards |
Re: Name of the source module without read rights Therefore your trying to use dxl to read module name that are hidden to you is apparently the wrong approach. You need to go to the admin and get permissions. However, if you are just currious about the name -- you can look at link sets to try and determine the module name. |
Re: Name of the source module without read rights Ron_Lewis - Thu Aug 13 09:18:04 EDT 2009 AAI Services, Textron dpechacek@aai.textron.com David.Pechacek@gmail.com |