You might consider changing how you store data just a little bit.
Instead of having two arrays, a single array with arrays as objects.
For example, you currently have:
Array 1 Array 2
--------- -----------
| | | |
--------- -----------
--------- -----------
| | | |
--------- -----------
--------- -----------
| | | |
--------- -----------
--------- -----------
| | | |
--------- -----------
And you want the order of the two to remain the same. I would propose
what you really want is:
Container array
----------------------------------------
| |
| --------- ----------- |
| | | ---------| | | objectAtIndex:0 is
an array of two items and so on.
| --------- ----------- |
| --------- ----------- |
| | | ---------| | |
| --------- ----------- |
| --------- ----------- |
| | | ---------| | |
| --------- ----------- |
| --------- ----------- |
| | | ---------| | |
| --------- ----------- |
----------------------------------------
And then you sort (using sortUsingSelector:) the Container Array.
Your selector would be within a category of NSMutalbeArray and compare
what you want to compare for the sort.
To get to objects in Array, you would use [[containerArray
objectAtindex:index] objectAtIndex:0] VS [Array 1 objectAtIndex:index].
You could also model the items as a new class (perhaps a bit cleaner
and easier to maintain) but this might not be what you want.
Steven
On Jul 28, 2008, at 10:08 AM, Jeff LaMarche wrote:
Hey all...
I've got two arrays that are related, so that object 0 in one array
corresponds to object 0 in the other array. I'd like to implement a
sort based on the values in one array, but to reorder both array so
that the order of the two arrays stays in sync and that object x in
one array continues to correspond to the same thing as object x in
the other array after the sort.
I've got a general idea of how to implement the sort, but since I
know that there are a few places in the system where array pairs are
used like this, I wanted to see if there was any out-of-the-box way
of handling this scenario - I'd hate to reinvent the wheel if it's
not necessary. Does anyone know if there is any easy way to do this?
TIA,
Jeff
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