RE: Dictionary or Array?
RE: Dictionary or Array?
- Subject: RE: Dictionary or Array?
- From: "Jonathan E. Jackel" <email@hidden>
- Date: Fri, 13 Jun 2003 10:24:06 -0400
Lorenzo:
Rethinking how your data model works may help. One option is to use a
setter for your main array that also updates your array of names. Something
like (warning: written in mail):
(void)addDictionaryToDatabase:(NSDictionary *)dict
{
[databaseArray addObject:dict];
[nameArray addObject:[dict objectForKey:name]];
}
You could then re-sort nameArray. It may be more efficient to find the
right place to put the new name and use insertObject:atIndex: instead.
Removing an entry from the database works in reverse:
(void)removeDictionaryFromDatabase:(NSDictionary *)dict
{
[nameArray removeObject:[dict objectForKey:name]];
[databaseArray removeObject:dict];
}
Jonathan
>
-----Original Message-----
>
From: email@hidden
>
[mailto:email@hidden]On Behalf Of Lorenzo
>
Sent: Friday, June 13, 2003 8:50 AM
>
To: email@hidden
>
Subject: Dictionary or Array?
>
>
>
Hi list,
>
I have a small database I created using a NSMutableArray of
>
NSMutableDictionary records.
>
>
e.g. If I need to get the field "name" of the record number 2 I do:
>
>
name = [[databaseArray objectAtIndex:2] objectForKey:@"name"];
>
>
Now I would like to get a pointer to the array containing all the entries
>
for the field "name" in order to get the list of all the "names" in my
>
database;
>
>
Currently I iterate through the databaseArray and get one by one all the
>
"name" entries. This requires time. And up above it's not dynamic
>
because if
>
the user adds a record or modifies a record, I have to rebuild the "names"
>
array list.
>
>
Is there a way to get "a pointer" to the vertical "names" array?
>
>
I thought to rebuild the database using a vertical NSMutableArray for each
>
field. Then in order to get a record I should iterate through the fields.
>
>
name = [namesArray objectAtIndex:2];
>
surname = [surnamesArray objectAtIndex:2];
>
...
>
>
Is it a good idea?
>
>
>
Best Regards
>
--
>
Lorenzo
>
email: email@hidden
>
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