使用属性的名称动态访问Objective-C属性

I know the string name of a property of an object. How would I go about getting and setting that property using the string?

我知道对象属性的字符串名称。我将如何使用字符串获取和设置该属性？

2 个解决方案

#1

While @weichsel is correct, there is a better way.

虽然@weichsel是正确的，但还有更好的方法。

Use:

使用：

[anObject valueForKey: @"propertyName"];

and

和

[anObject setValue:value forKey:@"propertyName"];

Obviously, @"propertyName" can be an NSString that is dynamically composed at runtime.

显然，@“propertyName”可以是在运行时动态组合的NSString。

This technique is called Key Value Coding and is fundamental to Cocoa.

这种技术称为键值编码，是Cocoa的基础。

Why this is better is because -valueForKey will do what is necessary to "box" whatever type the property returns into an object. Thus, if the property is of type int, it'll return an NSNumber instance containing the int.

为什么这样做更好是因为-valueForKey将执行“绑定”属性返回到对象的任何类型所需的操作。因此，如果属性是int类型，它将返回包含int的NSNumber实例。

This is much easier to deal with -- performSelector will only work for types that happen to fit into a pointer's worth of memory.

这更容易处理 - performSelector仅适用于恰好适合指针内存的类型。

Note that there is also -setValue:forKey:.

请注意，还有-setValue：forKey：。

#2

@synthesize propertyName automates the generation of getter and setter methods.

@synthesize propertyName自动生成getter和setter方法。

The compiler generates

编译器生成

- (id)propertyName;
- （id）propertyName;
- (void)setPropertyName;
- （void）setPropertyName;

If you have a selector as NSString, you can use performSelector:NSSelectorFromString.
e.g.:
[object performSelector:NSSelectorFromString(@"propertyName") ...]

如果您有一个选择器作为NSString，您可以使用performSelector：NSSelectorFromString。例如：[object performSelector：NSSelectorFromString（@“propertyName”）...]

#1