iOS中JSONModel的使用
Adding JSONModel to your project (https://github.com/icanzilb/JSONModel)
添加JSONModel到你的项目中
Requirements
需要的环境
ARC only; iOS 5.0+ / OSX 10.7+
SystemConfiguration.framework(需要导入系统库)
Get it as: 1) source files
Download the JSONModel repository as azip fileor clone it
下载JSONModel.zip文件
Copy the JSONModel sub-folder into your Xcode project
将它拷贝到你的项目中
Link your app to SystemConfiguration.framework
导入SystemConfiguration.framework框架
or 2) via Cocoa pods
使用Cocoa pods引入
In your project'sPodfileadd the JSONModel pod:
pod 'JSONModel'
If you want to read more about CocoaPods, have a look atthis short tutorial.
如果你想关于CocoaPods了解更多，请参考这个简单的教程
Source code documentation
源码文档
The source code includes class docs, which you can build yourself and import into Xcode:
源码包含了类的文档，你也可以自己编译并且导入到xcode
If you don't already haveappledocinstalled, install it withhomebrewby typing brew install appledoc.
如果你还没有安装appledoc，先安装appledoc
Install the documentation into Xcode by typing appledoc .
in the root directory of the repository.
在xcode上安装appledoc文档，在根目录下
Restart Xcode if it's already running.
重启xcode
Basic usage
基本使用
Consider you have a JSON like this:
假如你的JSON数据像这样：
{"id":"10", "country":"Germany", "dialCode": 49, "isInEurope":true}
Create a new Objective-C class for your data model and make it inherit the JSONModel class.
创建一个Objective-C类，继承自JSONModel
Declare properties in your header file with the name of the JSON keys:
将JSON中的keys在.h文件中声明为属性
#import"JSONModel.h"@interfaceCountryModel:JSONModel@property(assign,nonatomic)intid;@property(strong,nonatomic)NSString* country;@property(strong,nonatomic)NSString* dialCode;@property(assign,nonatomic)BOOLisInEurope;@end
There's no need to do anything in the .m file.
在.m文件中不需要做任何事情
Initialize your model with data:
用数据初始化你的model
#import"CountryModel.h"...NSString* json = (fetch here JSON from Internet) ...NSError* err =nil;CountryModel* country = [[CountryModel alloc] initWithString:json error:&err];
If the validation of the JSON passes you have all the corresponding properties in your model populated from the JSON. JSONModel will
also try to convert as much data to the types you expect, in the example above it will:
如果传过来的JSON合法，你所定义的所有的属性都会与该JSON的值想对应，甚至JSONModel会尝试去转换数据为你期望的类型，如上所示：
convert "id" from string (in the JSON) to an int for your class
just copy country's value
转换id，从字符串转换为int
convert dialCode from number (in the JSON) to an NSString value
转换diaCode，从number转换为字符串
finally convert isInEurope to a BOOL for your BOOL property
最后一个是将isInEurope转换为BOOL属性
And the good news is all you had to do is define the properties and their expected types.
所以，你所需要做的就是将你的属性定义为期望的类型
Online tutorials
在线教程
Official website:http://www.jsonmodel.com
Class docs online:http://jsonmodel.com/docs/
Step-by-step tutorials:
傻瓜教程：
How to fetch and parse JSON by using data models
Performance optimisation for working with JSON feeds via JSONModel
How to make a YouTube app using MGBox and JSONModel
Examples
例子
命名自动匹配
{
"id": "123",
"name": "Product name",
"price": 12.95
}
@interfaceProductModel:JSONModel@property(assign,nonatomic)intid;@property(strong,nonatomic)NSString* name;@property(assign,nonatomic)floatprice;@end@implementationProductModel@end
文／jueyingxx（简书作者）
原文链接：http://www.jianshu.com/p/3cce56f374b4
著作权归作者所有，转载请联系作者获得授权，并标注“简书作者”。
