我想将部分标题显示为一月、二月、三月(根据数据来自 api),而不是显示一月、三月、二月。
我从后端 api 获取这些数据。
wholeDic : NSDictionary = {
January = (
{
date = "20-jan-18";
subtitle = "Only four days remaining";
title = "School fees of August, 2018School fees of August, 2018";
},
{
date = "21-jan-18";
subtitle = Holi;
title = "Holiday on third march";
}
);
february = (
{
date = "20-feb-18";
subtitle = "Paricipate in this activity";
title = "Annual function will be held in feb,2018";
},
{
date = "20-12-18";
subtitle = "Holiday issue by Govt. of India";
title = "Bharat Band";
}
);
march = (
{
date = "20-feb-18";
subtitle = "Paricipate in this activity";
title = "Annual function will be held in feb,2018";
},
{
date = "20-feb-18";
subtitle = "Paricipate in this activity";
title = "Annual function will be held in feb,2018";
}
);}
现在我已经获取“密钥”并添加到数组中
for (key, value) in wholeDic{
sectionTitleArray.add(key)
print(sectionTitleArray)
}
当我打印 sectionTitleArray 时,控制台显示January, March, February
而不是显示January, February, March
.
我知道 Dictionary 是一个无序集合,但我想知道如何按顺序获取键?
我的 UitableView 数据源和委托是
func numberOfSections(in tableView: UITableView) -> Int{
return sectionTitleArray.count
}
func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
let sectionTitle : String = sectionTitleArray.object(at: section) as! String
let sectiondes:NSArray = wholeDic.object(forKey: sectionTitle) as! NSArray
return sectiondes.count
}
这是我的 tableView 。一切正常,但我想根据 api 数据显示像 JANUARY、FEBRUARY、MARCH 这样的月份。
func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
let cellIdentifier = "cell"
let cell = tableView.dequeueReusableCell(withIdentifier: cellIdentifier, for: indexPath) as! frameNboundTableViewCell
let sectionTitle : String = sectionTitleArray.object(at: indexPath.section) as! String
let contentArr : NSArray = wholeDic.object(forKey: sectionTitle) as! NSArray
let contentDic : NSDictionary = contentArr.object(at: indexPath.row) as! NSDictionary
print(contentDic)
cell.titleLbl.text = contentDic.value(forKey: "title") as? String
cell.titleLbl.numberOfLines = 0
return cell
}