从以前的字典数组创建和排序字典数组答案

【问题标题】：Create and sort array of dictionaries from previous array of dictionaries从以前的字典数组创建和排序字典数组
【发布时间】：2014-06-06 10:04:44
【问题描述】：

我真的试着不在这里问这个问题，但经过几个小时的尝试，我被困住了。所以这里......希望我能清楚。

我想要的输出是这样的：

({
    following = "";
    followingMe = "";
},{
    following = "";
    followingMe = "";
},{...})

进来的数据是这样的，我要做的就是排序

({
    user = "me";
    following = "sam";
},{
    user = "sam";
    following = "me";
},{
    user = "foo";
    following = "me";
},{
    user = "me";
    following = "bar";
},...)

我想整理它，所以如果我关注“sam”而他也在关注我，那么它就是一本字典。如果 "foo" 跟着我，那么 following="(null)" 和 followingMe="sam"...等等。这就是为什么我不想问这个问题...很难解释。

这是我拥有的代码，但它只是一遍又一遍地给我最后一个条目......错字是什么？

NSMutableDictionary *temp = [NSMutableDictionary new];
int i = 0;
while (i < array.count) {
    [temp removeAllObjects];
    if ([[[array objectAtIndex:i]valueForKey:@"user"]isEqualToString:_user]){
    //means I'm following this person
        [temp setValue:[[array objectAtIndex:i] valueForKey:@"following"] forKey:@"following"];

        if ([[_allfriends valueForKey:@"followingMe"]containsObject:[[array objectAtIndex:i] valueForKey:@"following"]]) {
        //check to see if this person is already following me

            [[_allfriends objectAtIndex:[[_allfriends valueForKey:@"followingMe"] indexOfObject:
            [[array objectAtIndex:i] valueForKey:@"following"]]] setObject:[[array objectAtIndex:i] valueForKey:@"following"] forKey:@"following"];
            //trying to update at the index where this person is following me

        }else{
            [temp setObject:@"(null)" forKey:@"followingMe"];
            [_allfriends addObject:temp];
        }
    }else{
        //this person is following me
        [temp setObject:[[array objectAtIndex:i]valueForKey:@"user"] forKey:@"followingMe"];

        if ([[_allfriends valueForKey:@"following"]containsObject:[[array objectAtIndex:i] valueForKey:@"user"]]) {
        //check to see if I'm following this pseron

            [[_allfriends objectAtIndex:[[_allfriends valueForKey:@"following"] indexOfObject:
            [[array objectAtIndex:i] valueForKey:@"user"]]] setObject:[[array objectAtIndex:i] valueForKey:@"user"] forKey:@"followingMe"];
            //try to update followingMe at that index

        }else{
            [temp setObject:@"(null)" forKey:@"following"];
            [_allfriends addObject:temp];
        }
    }
    i++;
}

所以逻辑是我正在检查我的名字在哪里（user 或following）。然后我正在检查这个人是否已经是followingMe。它是_allfriends 中的一个条目，我需要获取该索引并且我想更新“following”...有意义吗？我在这里做错了什么？

【问题讨论】：

标签： ios arrays dictionary

【解决方案1】：

你可以用 Set 对数组进行排序

SortedArr = [NSMutableArray arrayWithArray:[[NSSet setWithArray: UnsortedArr] allObjects]];

【讨论】：

我不确定这对我有什么帮助，但它只是给了我原始数组
抱歉，但这会从数组中删除重复元素...不排序

【解决方案2】：

显然问题出在NSMutableDictionary *temp = [NSMutableDictionary new]; 上。我不确定this post 是否适用，但它让我重写了我的代码。 temp 值附加到 _allfriends 数组，所以当 temp 改变时，_allfriends 中的一切都改变了。我的循环逻辑是正确的，但我改写成这样：

NSString *key = [NSString new];
for (int i = 0; i < array.count; i++) {
    if ([[[array objectAtIndex:i]valueForKey:@"user"]isEqualToString:_user]){
        key = [[array objectAtIndex:i] valueForKey:@"following"];
        if ([[_allfriends valueForKey:@"followingMe"]containsObject:key]) {
            [_allfriends removeObjectAtIndex:[[_allfriends valueForKey:@"followingMe"] indexOfObject:key]];
            [_allfriends insertObject:[NSDictionary dictionaryWithObjectsAndKeys:key, @"following", key, @"followingMe", nil] atIndex:_allfriends.count];
        }else{
            [_allfriends insertObject:[NSDictionary dictionaryWithObjectsAndKeys:key, @"following", @"(null)", @"followingMe", nil] atIndex:_allfriends.count];
        }
    }else{
        key = [[array objectAtIndex:i] valueForKey:@"user"];
        if ([[_allfriends valueForKey:@"following"]containsObject:key]) {
            [_allfriends removeObjectAtIndex:[[_allfriends valueForKey:@"following"] indexOfObject:key]];
            [_allfriends insertObject:[NSDictionary dictionaryWithObjectsAndKeys:key, @"following", key, @"followingMe", nil] atIndex:_allfriends.count];
        }else{
            [_allfriends insertObject:[NSDictionary dictionaryWithObjectsAndKeys:key, @"followingMe", @"(null)", @"following", nil] atIndex:_allfriends.count];
        }
    }
}

希望对某人有所帮助...

【讨论】：