I've been trying to find a code for ten days to process only the nodes that are selected and for more laps I've taken, I'm not able to find the solution. The most I have come to know is that node is totally or partially selected (csChecked, or csMixed states), but then I get stuck and, for example, in node Banco (no domiciliado), I don't know how to process only the marked nodes, in this case it would be Manzanares. Can you give me some code that gives me a hint to do it?
If I understand your situation you do not care about csMixed, only csChecked. In your example the values for each 'parent' node are the total for the subnodes.
If this is the case, then you want to only check those nodes which
have no children
AND are checked.
So you would probably want code looking something like:
for int(nLoop=0; nLoop<pControl->Items->Count; nLoop++)
{
if( (!(pControl->Items->Item[nLoop]->HasChildren)) &&
(pControl->Items->Item[nLoop]->State==csChecked) )
{
// call your process routine
}
}
I don't have any reference for the control you are trying to use so I don't know the properties on it, but I have guessed what you need. You should be able to find your way from here if I have understood what you want to do correctly.
Related
For my program, I have multiple tables inside of each other for organization and readability. These tables can look something like this
local buttons = {
loadingScreen = {
bLoadingEnter = function()
end
}, ...
}
What I want to do is find the first element named bLoadingEnter in the table. I dont know that the element named bLoadingEnter will be in loadingScreen. I've thought of getting all the keys in the table then check them. I couldn't get that to work. Any help would be appreciated!
If execution speed is in any way relevant, the answer is:
you don't
Looking up a key in a nested table takes a lot longer than looking it up in just a normal table. The next problem is that of uniqueness. Two or more nested tables can have the same key with different values, which may lead to odd bugs down the road. You'd either have to check for that when inserting (making your code even slower) or just hope things miraculously go well and nothing explodes later on.
I'd say just use a flat table. If your keys are well named (like bLoadingEnter) you'll be able to infer meaning from the name either way, no nesting required.
That being said, nested tables can be a good option, if most of the time you know what path to take, or when you have some sort of ordered structure like a binary search tree. Or if speed really isn't an important factor to consider.
Okay try
for _, item in ipairs(buttons) do
if item.bLoadingEnter ~= nil then
return item.bLoadingEnter
end
end
I want to mark node as deleted (and not actually delete the node) and i'm not sure if to set a property in the node(deleted:0/1) or set label to the deleted node. which way is more efficient and right?
From my point of view it's label, because they are indexed by default.
This article could help you - http://graphaware.com/neo4j/2015/01/16/neo4j-graph-model-design-labels-versus-indexed-properties.html
It depends entirely on your use case. In many cases you will actually delete the node. For others you can set a property or add a label. From a performance point of view there shouldn't be much of a difference.
The most important thing is to understand how your application is going to interact with the node that is marked deleted. Do you still want it searchable? How are people searching now and if you don't want it searchable what is the easiest way for you to modify your query to exclude deleted information? Will you ever need to restore the node or query it at a later time?
I am looking for a template of sorts for merging two linked chains that have already been sorted. I'm still fairly new to Java, and this seems to be a pretty challenging task to accomplish with the limited knowledge I have. I have an understanding of how to merge sort an array, but when it comes to linked lists I seem to be drawing blanks. Any help you all could give me, be it actual code or simply advise on where to start, would be greatly appreciated.
Thank you for your time!
If the two linked list are already sorted, then it is so easy to merge those two together. I am gonna tell you the algorithm but you need to write the code yourself since it seems like a school project. First you make a new linked list, and then assign the head of the new list to be the min of list1Head and list2Head, then you just walk the two list, each time picking the min of the current node of the two list and append to the new created list, make the current to be .Next if it got picked. If one of the list doesn't have more nodes, then append the rest of another list directly to the new list. Done
Can't you look at the first element in each list and take the smallest. This is the start of the new list. Remove this from the front ofwhichever list it came from. Now look at the first element again and take the smallest and make it the second element in the new list. Then just repeat this process zipping the two lists together.
If you want to avoid creating a new list the just find the smallest then look at the thing is pointing at and the beginning of the other list and see which is smaller. If you are not already pointing at the smaller one the update the pointer so it is. Then rinse and repeat.
Using Neo4j 1.9.3 -
I want to create a music program listing. On a given program there may be three pieces being performed. Each piece has a composer associated with them, and may appear on many different programs, so I can't put sequence numbers on the piece nodes.
I assume I can create the program, with relationships to each piece like so:
(program1)-[:PROGRAM_PIECE {program_seq: 1}]->(piece1)
(program1)-[:PROGRAM_PIECE {program_seq: 2}]->(piece2)
(program1)-[:PROGRAM_PIECE {program_seq: 3}]->(piece3)
My question is, how do I query the graph so that the pieces are in order of the relationship property program_seq? I'm fine using ORDER BY with node properties, but have not been successful with relationships (story of my life...)
If you like it, lock it down: that is, bind it to a variable. Then you can use ORDER BY the same way you would with node properties. If you have retrieved your program as (program1) you can do something like
MATCH (program1)-[r:PROGRAM_PIECE]->(piece1)
RETURN program1, r, piece1
ORDER BY r.program_seq
I have done the same thing recently to keep track of chess moves in a particular game. It is the same thing as node properties.
start program = node(*) // or better yet, use a real index query to find the program
match (program)-[program_piece:PROGRAM_PIECE]->(piece)
return program, piece
order by program_piece.program_seq
I'm working on understanding how to use linked lists to improve performance and create activity feeds on Neo4j.. Still working on learning Cypher, so I have a question.. I've found some examples of linked lists, but I need lists with bigger examples to finally put all the pieces together in my head..
I've used this code from grepcode and have found it to be more helpful than the example in the Neo4j manual. Yet I'm still a bit confused.. Can someone modify it to have say seven nodes with seven items in the linked list, and then insert a node on the front of it?
Yea, I'm trying to put the latest status update on the top of the linked list. This example doesn't really do that, but it's close.. so looking for some mods.. No, I'm not really coding yet, still trying to master Cypher first - will continue to study it for the next two weeks... Have the Ruby on Rails side working .. just need to understand linked lists used with Cypher/Neo a bit better.
CREATE zero={name:0,value:0}, two={value:2,name:2}, zero-[:LINK]->two-[:LINK]->zero
==== zero ====
MATCH zero-[:LINK*0..]->before,
after-[:LINK*0..]->zero,
before-[old:LINK]->after
WHERE before.value? <= 1 AND
1 <= after.value?
CREATE newValue={name:1,value : 1},
before-[:LINK]->newValue,
newValue-[:LINK]->after
DELETE old
==== zero ====
MATCH p = zero-[:LINK*1..]->zero
RETURN length(p) as list_length
What I'm trying to do in my mind is understand the before after and zero data sets - I almost have it, but want to see how it's done on a set with more than two starting nodes so as to clear up any confusion
Thank you!
The node in front is special as it doesn't have a incoming link relationship. Usually you also keep the connection to the head node somewhere, so this is about replacing this link to the head node and moving the head node one step further away. Something like this:
start user=node:node_auto_index(user="me")
match user-[old:MESSAGES]->head
delete old
create new_heads = { title: "Title", date : 2348972389, text: "Text" },
user-[:MESSAGES]->new_head-[:LINK]->head