Object pointer issue

When I'm done making changes, I replace the added record in the database

If you're changing records that are already in the array, there's no need to replace anything. Do this:

DatabaseRecord *currentRecord;

currentRecord = [database objectAtIndex:someIndex];
currentRecord.name = @"tom";
currentRecord.location = @"los angeles";

currentRecord = [database objectAtIndex:someOtherIndex];
currentRecord.name = @"harry";
currentRecord.location = @"new york";

Now you've just changed the contents of two of the records in the database -- no need to replace anything. The reason you can do this is that currentRecord is just a pointer to an object that already exists.

If you want to modify every record in the array, you can do:

for (currentRecord in database) {
    // make changes to currentRecord here
    // no need to replace the record in the array -- it's already there!
}

When you're adding new records to the database, on the other hand, you'll need to make sure that you create new objects each time. You can do that by copying an existing object, or by instantiating a new object each time yourself:

for (int i = 0; i < 10; i++) {
    currentRecord = [[DatabaseRecord alloc] init];
    [database addObject:currentRecord];
    // add [currentRecord release]; for non-ARC builds
}

or:

currentRecord = [[DatabaseRecord alloc] init];
for (int i = 0; i < 10; i++) {
    [database addObject:[currentRecord copy]];
    // make the previous line [database addObject: [[currentRecord copy] autorelease]]; for non-ARC builds
}

If you're not using ARC, you'll need to release the objects you create appropriately, as shown in the comments above.

I'm sure there's an easy fix to this, such as not passing a pointer

Nope. The only way to refer to an object in Objective-C is via pointer. The fix isn't difficult, though -- you just need to understand how pointers work a little better.