Db4o updating arraylist problem

Then it modifies the loaded object structure in the run-time memory.

Lots of enhancements in the areas of Java FX, php, ruby, OSGi ….

How Declarative Updating works (Slides as a PDF): Here The content is also in ar Xiv: Here Download Java demo Implementation of the embed method as a Zip file: Here The demo implementation in Git Hub: Here In this article I introduce a new communication theory for complex information represented as a direct graph of nodes.

One embed method call can replace dozens of lines of complicated updating code in a traditional client program of an object database, which is a huge improvement.

As a declarative method the embed method takes only one natural parameter, the root object of a modified object structure in the run-time memory, which makes it extremely easy to use.

The single natural parameter for the method would be the root object of a modified object structure because it naturally defines the modified object structure in the run-time memory.

The modified object structure could also be multi-rooted, but at the moment we can ignore it.From a Java program the embed method is called in the following way: db.embed(s); where s is the root object of a modified object structure in the run-time memory.The communication theory behind the embed method states that modified complex information represented as a directed graph of nodes can always be transferred back to its original system in an exact and meaningful way.i have lot of trouble trying figure out.i have seen lots of people show complex (complex me) ways of fixing these columns have no idea how in this, using way, here's code: elif menu==2: print("name=================phone number===") x in book.keys(): s=1 p=s 1 print(str(p) ")",x,"\t\t",book[x]) s=p also if there else doing more efficient, please tell me new python., can give more code if needed.sorry, didn't explain well, output want like:name===================phone number john 0425111222 alexander 81238211 firstand lastname 99938880 thanks, leo you can that:i = 1 print(' '.format('#', 'name', 'number')) name, number in book.items(): print(' '.format(i, name, number)) = 1 it simple , example output:# name number 1 contact1 555 2 contact2 …But if object structures reside in a remote memory system, like in a persistent store, in an object database, manipulating or updating them is a much more difficult task, especially in complex cases.