ios NSPredicate 与整数比较
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NSPredicate compare with Integer
提问by wolfrevo
How can I compare two NSNumbers or a NSNumber and an Integer in an NSPredicate?
如何在 NSPredicate 中比较两个 NSNumber 或一个 NSNumber 和一个整数？
I tried:
我试过：
NSNumber *stdUserNumber = [NSNumber numberWithInteger:[[NSUserDefaults standardUserDefaults] integerForKey:@"standardUser"]];
fetchRequest.predicate = [NSPredicate predicateWithFormat:@"userID == %@", stdUserNumber];
Well, this doesn't work ... anyone has an idea how to do it? I'm sure it's pretty easy but I wasn't able to find anything ...
好吧，这不起作用……有人知道该怎么做吗？我确定这很容易，但我找不到任何东西......
回答by EmptyStack
NSNumberis an object type. Unlike NSString, the actual value of NSNumberis not substitued when used with %@format. You have to get the actual value using the predefined methods, like intValuewhich returns the integer value. And use the format substituer as %das we are going to substitute an integer value.
NSNumber是一个对象类型。与 不同NSString，NSNumber的实际值在与%@format 一起使用时不会被替换。您必须使用预定义的方法获取实际值，例如intValue返回整数值。并使用格式替换器，%d因为我们将替换一个整数值。
The predicate should be,
谓词应该是，
predicateWithFormat:@"userID == %d", [stdUserNumber intValue]];
回答by Josip B.
If you are using predicateWithSubstitutionVariables: and value you want to predicate is NSNumber with integer value then you need to use NSNumber in predicate instead of NSString as substitution value. Take a look at example:
如果您使用 predicateWithSubstitutionVariables: 并且您想要谓词的值是 NSNumber 和整数值，那么您需要在谓词中使用 NSNumber 而不是 NSString 作为替换值。看一个例子：
NSPredicate *genrePredicate = [NSPredicate predicateWithFormat:[NSString stringWithFormat:@"genreID == $GENRE_ID"]]; //if we make only one NSPredicate if would look something like this: [NSPredicate predicateWithFormat:@"genreID == %d", [someString integerValue]];
NSPredicate *localPredicate = nil;
for(NSDictionary *genre in genresArray){
localPredicate = [genrePredicate predicateWithSubstitutionVariables:@{@"GENRE_ID" : [NSNumber numberWithInteger:[genre[@"genre_id"] integerValue]]}];
}
