在Swift中使用谓词 [英] Using Predicate in Swift
问题描述
我正在学习这个教程(学习Swift)我的第一个应用程序:
http ://www.appcoda.com/search-bar-tutorial-ios7/
I'm working through the tutorial here (learning Swift) for my first app: http://www.appcoda.com/search-bar-tutorial-ios7/
我坚持这一部分(Objective-C代码):
I'm stuck on this part (Objective-C code):
- (void)filterContentForSearchText:(NSString*)searchText scope:(NSString*)scope
{
NSPredicate *resultPredicate = [NSPredicate predicateWithFormat:@"name contains[c] %@", searchText];
searchResults = [recipes filteredArrayUsingPredicate:resultPredicate];
}
有人可以建议如何在Swift中为NSPredicate创建一个等价物吗?
Can anyone advise how to create an equivalent for NSPredicate in Swift?
推荐答案
这实际上只是一个语法切换。好的,所以我们有这个方法调用:
This is really just a syntax switch. OK, so we have this method call:
[NSPredicate predicateWithFormat:@"name contains[c] %@", searchText];
在Swift中,构造函数跳过blahWith ...部分,只使用类名作为函数,然后直接进入参数,所以 [NSPredicate predicateWithFormat:...]
将变为 NSPredicate(格式:...)
。 (另一个例子, [NSArray arrayWithObject:...]
将成为 NSArray(对象:...)
。这是一个常规的Swift中的模式。)
In Swift, constructors skip the "blahWith…" part and just use the class name as a function and then go straight to the arguments, so [NSPredicate predicateWithFormat: …]
would become NSPredicate(format: …)
. (For another example, [NSArray arrayWithObject: …]
would become NSArray(object: …)
. This is a regular pattern in Swift.)
所以现在我们只需要将参数传递给构造函数。在Objective-C中,NSString文字看起来像 @
,但在Swift中我们只使用字符串的引号。所以这给了我们:
So now we just need to pass the arguments to the constructor. In Objective-C, NSString literals look like @""
, but in Swift we just use quotation marks for strings. So that gives us:
let resultPredicate = NSPredicate(format: "name contains[c] %@", searchText)
事实上,这正是我们需要的。
And in fact that is exactly what we need here.
(顺便提一下,你会注意到其他一些答案,而是使用像这样的格式字符串name contains [c] \(searchText)
。这不是这是使用字符串插值,这与谓词格式不同,通常不适用于此。)
(Incidentally, you'll notice some of the other answers instead use a format string like "name contains[c] \(searchText)"
. That is not correct. That uses string interpolation, which is different from predicate formatting and will generally not work for this.)
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