Re: [libreoffice-users] Re: Save UNO state?

Hi,
I wrote some code some time ago. The code also checks the differences 
between two states of an object.
The first part is similar to yours, it's also a recursive function, but 
it creates a dict and you can use a level to stop searching, so you 
won't get an infinite cycle.


The second part - I think, I used some code from stackoverflow - 
compares two nested dicts with each other and prints out the results.

It works best inside of an extension, where you can keep objects alive.

so it would be:
level = 2
self.res1 = get_attribs(object,level )

#do something
# and start the function again with:

self.res2 = get_attribs(object,level )
findDiff(self.res1, self.res2)

Regards,
Xaver


Here's the code:



ctx = uno.getComponentContext()
smgr = ctx.ServiceManager
desktop = smgr.createInstanceWithContext( "com.sun.star.frame.Desktop",ctx)
doc = desktop.getCurrentComponent()
current_Contr = doc.CurrentController
viewcursor = current_Contr.ViewCursor

object = doc
max_lvl = 3

def get_attribs(obj,lvl):
    results = {}
    for key in dir(obj):

        try:
            value = getattr(obj, key)
            if 'callable' in str(type(value)):
                continue
        except :
            #print(key)
            continue

        if key not in results:
            if type(value) in (
                               type(None),
                               type(True),
                               type(1),
                               type(.1),
                               type('string'),
                               type(()),
                               type([]),
                               type(b''),
                               type(r''),
                               type(u'')
                               ):
                results.update({key: value})

            elif lvl < max_lvl:
                try:
                    results.update({key: get_attribs(value,lvl+1)})
                except:
                    pass

    return results


diff = []

def findDiff(d1, d2, path=""):
    for k in d1.keys():
        if not d2.has_key(k):
            print path, ":"
            print k + " as key not in d2", "\n"
        else:
            if type(d1[k]) is dict:
                if path == "":
                    path = k
                else:
                    path = path + "->" + k
                findDiff(d1[k],d2[k], path)
            else:
                if d1[k] != d2[k]:
                    diff.append((path,k,d1[k],d2[k]))
                    path = ''


res1 = get_attribs(object,1)

viewcursor.gotoEnd(False)
viewcursor.setString('Test ')

res2 = get_attribs(object,1)

findDiff(res1, res2)

for d in diff:
    print(d)
    time.sleep(.4)







Am 22.08.2015 um 20:24 schrieb Piet van Oostrum:
> Hi-Angel wrote:
>
>   > Okay, I wrote a python code that explores every property, and doesn't
>   > fall into an infinite cycle. It is
>   > [...]
>   > However it produces too much output, e.g. from a test document it
>   > produced 1gb of output. I think the problem is that most elements
>   > still appears in output many times — the check that in backtrace
>   > wasn't the current element is ensures only that it wouldn't fall in an
>   > infinite cycle.
>
> You can get an infinite cycle at any level, so only checking at the top level is not good enough. 
> You will have to check at each node.


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