我有一个NSMutableArray
和一个NSArray
。两者都由NSDictionarys
本身组成。 两个样本结构如下:根据字典中的参数合并两个由字典组成的NSArrays
NSMutableArray
[
{
objectId = 4274;
name = orange;
price = 45;
status = approved;
},
{
objectId = 9035;
name = apple;
price = 56;
status = approved;
},
{
objectId = 7336;
name = banana;
price = 48;
status = approved;
}
.
.
.
.
]
和NSAraay是
NSArray
[
{
objectId = 4274;
name = orange;
price = 106;
status = not_approved;
},
{
objectId = 5503;
name = apple;
price = 56;
status = approved;
}
]
我想是合并这两个数组,这样,如果在NSArray
任一元素的任何元素相同objectId
在NSMutableArray
中,NSArray
中的元素应覆盖NSMutableArray
中的元素。
因此,在这种情况下,最终的合并数组应该是这样的
MergedArray
[
{
objectId = 4274;
name = orange;
price = 106;
status = not_approved;
},
{
objectId = 9035;
name = apple;
price = 56;
status = approved;
},
{
objectId = 7336;
name = banana;
price = 48;
status = approved;
},
{
objectId = 5503;
name = apple;
price = 56;
status = approved;
}
.
.
.
.
]
只有这样,这个我知道的是,通过两个阵列迭代和合并。有没有更好的方法?任何帮助将不胜感激。
编辑:
继dasblinkenlights建议,我做了以下方式
-(NSMutableArray*)mergeTwoArray:(NSArray*)array1 :(NSArray*)array2
{
//array1 will overwrite on array2
NSSet* parentSet = [NSSet setWithArray:array2];
NSMutableDictionary *dict = [NSMutableDictionary dictionary];
for (NSDictionary *item in parentSet)
[dict setObject: item forKey: [item objectForKey:@"objectId"]];
NSLog(@"initial dictionary is %@",dict);
for (NSDictionary *item in array1)
[dict setObject: item forKey: [item objectForKey:@"objectId"]];
NSLog(@"final dictionary is %@ with all values %@", dict,[dict allValues]);
return [NSMutableArray arrayWithArray:[dict allValues]];
}
#9035在您的合并数组中出现两次。那是故意的吗? – 2012-04-05 14:48:03
不.. ..!我将编辑问题。感谢您指出 – chatur 2012-04-05 14:52:48