[英]Filter NSMutableArray with All the values of nsarray
我有這樣的MutableArray ...
restaurants = ({
cuisines = ({
"cuisine_id" = 1;
"cuisine_name" = Indian;
},
{
"cuisine_id" = 2;
"cuisine_name" = SriLanka;
}
);}
{ cuisines = ({
"cuisine_id" = 1;
"cuisine_name" = Chinese;
},
{
"cuisine_id" = 2;
"cuisine_name" = Afghani;
}
);}
{cuisines =({
"cuisine_id" = 1;
"cuisine_name" = Chinese;
},
{
"cuisine_id" = 2;
"cuisine_name" = Afghani;
}
);}
)
而且我正在過濾包含有Cuisine_name多個值的Restaurants Array。例如,如果我有一個,
像這樣的Nsarray cuisine_name(印度,斯里蘭卡,中國)
我想根據cusine_name的每個對象過濾我的Restaurants數組。
目前我正在這樣做。
NSPredicate *predicateCuisines = [NSPredicate predicateWithFormat:@"(cuisines.cuisine_name = %@)",[cuisineTypes valueForKey:@"cuisine_name"]];
[self.arrRestaurantList filterUsingPredicate:predicateCuisines];
但是它使用描述...進行過濾...對於包含整個三個值(印度,斯里蘭卡,中文)的“餐廳數組”,但是我想使用Cuisine_name的每個對象進行過濾。 任何幫助將不勝感激..
請使用這個
NSArray* arrtu4 = @[@{@"cuisines" : @[@{@"cuisine_id" : @1,@"cuisine_name" : @"Indian"},@{ @"cuisine_id" : @2, @"cuisine_name" : @"SriLanka"}]},@{ @"cuisines" : @[@{@"cuisine_id" : @1,@"cuisine_name" : @"Chinese"},@{@"cuisine_id" : @2,@"cuisine_name" : @"Afghani"}]},@{@"cuisines" :@[@{@"cuisine_id" :@1,@"cuisine_name" : @"Chinese"},@{ @"cuisine_id" : @2,@"cuisine_name" : @"Afghani"}]}];
NSMutableArray *arrryfilted = [NSMutableArray array];
NSArray *arrayCountrys = @[@"Chinese",@"SriLanka",@"Indian"];
for (int i = 0; i<arrtu4.count; i++)
{
NSArray *filterArray2 = [[arrtu4 objectAtIndex:i]valueForKey:@"cuisines"];
NSArray *filterArray3 = [filterArray2 filteredArrayUsingPredicate:[NSPredicate predicateWithBlock:^BOOL(id _Nonnull evaluatedObject, NSDictionary<NSString *,id> * _Nullable bindings) {
return [arrayCountrys containsObject:[evaluatedObject valueForKey:@"cuisine_name"]] ;
}]];
[arrryfilted addObjectsFromArray:filterArray3 ];
}
NSLog(@"%@",arrryfilted);
輸出:
2016-08-20 16:44:59.856 coredataReltaion[6762:152672] (
{
"cuisine_id" = 1;
"cuisine_name" = Indian;
},
{
"cuisine_id" = 2;
"cuisine_name" = SriLanka;
},
{
"cuisine_id" = 1;
"cuisine_name" = Chinese;
},
{
"cuisine_id" = 1;
"cuisine_name" = Chinese;
}
)
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