I would like to create a traceability column that shows for every objects in my current module (A) the linked objects in another module (B) |
Re: show descendants of an object
Bonjour,
Object o
Object po = current
for o in po do {
print (o."Object Heading") " is a child of "
print (po."Object Heading") "\n"
}
Object o = current
Object co = first o
if (null co) {
print "Current object has no children.\n"
} else {
if ((last o) == co) {
print "current has one child: " (o."Object
Heading") "\n"
print (identifier o) " == " (identifier
parent co) "\n"
}
}
if (null o[3])
print "current object does not have 3rd
child\n"
if (null previous o)
print "Current object is first in module.\n"
if (null next o)
print "Current object is last in module.\n"
if (!null next o) {
Object here = previous next o
print (identifier o) " and " (identifier
here) " are the same\n"
}
|
Re: show descendants of an object ViewDef defn = create(current Module, true/false) useDescendents(viewdef, true/false)) // true in your case What method are you using in order to find the link in module B? |
Re: show descendants of an object
It does not seem reasonable to me to presume that when A links to B, that it also links to all descendants of B. But lets suppose B is some sort of "as per the following list:" object, and has text children, one for each list element.
display some information about othero
Object otheroChild
for otheroChild in othero do
{ display some information about otheroChild
}
Buffer bResults = create()
if (obj is a 'clever' parent object) then
{ bResults = some info about clever parent obj
for oChild in obj do
{ bResults += "\n"
bResults += some info about child oChild
}
}
else
{ bResults = some info about normal object obj
}
obj.attrDXLName = tempStringOf(bResults)
delete(bReSults)
|