将NSDictionary条目添加到NSMutableDictionary

I have a for loop that goes through a series of dictionaries in an array.

我有一个for循环，通过一个数组中的一系列字典。

How can I consolidate all the dictionaries entries as it goes through the for loop into one NSMutableDictionary?

如何通过for循环将所有词典条目整合到一个NSMutableDictionary中？

I tried addEntriesFromDictionary, but its not working. Thanks for your help.

我试过addEntriesFromDictionary，但它无法正常工作。谢谢你的帮助。

for (int i=0; i<sections.count; i++){

    formElements    = [[sections objectAtIndex:i]objectForKey:@"Dictionary"];        
}

4 个解决方案

#1

NSMutableDictionary * mutableDict = [NSMutableDictionary dictionary];

for (NSDictionary * formElements in sections)
{
    [mutableDict addEntriesFromDictionary:formElements];
}

This should work if It's correct that they don't share any keys.

如果它们不共享任何键是正确的，这应该有效。

#2

you can add dictionary object like below.

你可以添加如下的字典对象。

NSMutableDictionary *mDict=[NSMutableDictionary dictionary];
    [mDict addEntriesFromDictionary:DictObj];

#3

NSMutableDictionary *mDict=[[NSMutableDictionary alloc]init];
    NSMutableDictionary *mDict2=[[NSMutableDictionary alloc]init];

//later suppose you have 5 object in mDict and 2 object in mDict2. combine in this Way.
    NSMutableArray *keys=[[NSMutableArray alloc]init];
    NSMutableArray *obj=[[NSMutableArray alloc]init];

    keys=[[mDict allKeys] mutableCopy];
    obj=[[mDict allValues] mutableCopy];

    [keys addObjectsFromArray:[mDict2 allKeys]];
    [obj addObjectsFromArray:[mDict2 allValues]];

    NSMutableDictionary *dict=[[NSMutableDictionary alloc]initWithObjects:obj forKeys:keys];

#4

You can enumerate your dictionaries with -objectEnumerator or other methodfs of NSDictionary.

您可以使用-objectEnumerator或NSDictionary的其他方法f枚举您的词典。

So inside your loop you enumerate your dictionary and add all objects an keys into one big dictionary.

因此，在循环中，您枚举字典并将所有对象添加到一个大字典中。

#1

NSMutableDictionary * mutableDict = [NSMutableDictionary dictionary];

for (NSDictionary * formElements in sections)
{
    [mutableDict addEntriesFromDictionary:formElements];
}