IOS Swift 从数组中搜索表 [英] IOS Swift Searching table from an Array
问题描述
我刚刚开始学习 swift,我正在研究 tableview 和 searchbar 功能.下面我有我的数组,它是水果列表:
var fruits: [[String]] = [["Apple", "Green"],["Pear", "Green"], ["Banana", "Yellow"], ["Orange", 橘子"]]
我将它们放在表格视图中,以水果的名称作为标题,以颜色作为副标题.我正在尝试使用搜索栏进行过滤,但我似乎无法正确使用.我只想搜索水果的名称而不是颜色.
varfilteredFruits = [String]()var shouldShowSearchResults = falsefunc searchBar(_ searchBar: UISearchBar, textDidChange searchText: String) {过滤的Fruits.removeAll()变量 i = 0当我 <水果数{varfilteredFruits =fruits[i].filter({ (fruit: String) -> Bool in返回fruit.lowercased().range(of: searchText.lowercased()) != nil})如果搜索文本 != ""{shouldShowSearchResults = true如果filteredItems.count >0{filteredFruits.append(filteredItems[0])过滤项目.removeAll()}}别的{shouldShowSearchResults = false}我 += 1}self.tableView.reloadData()}
我确实得到了返回的结果,但它混淆了字幕和标题,并且没有返回正确的结果.有人能指出我正确的方向吗?
我不明白你为什么使用某种 while 循环迭代结果.相反,我建议您利用以下功能:
func filterFruits(searchText: String) ->[[细绳]] {守卫搜索文本!="其他{回果}让针 = searchText.lowercased()返回fruits.filter {fruitObj in返回fruitObj.first!.lowercased().contains(needle)}}
该函数返回名称中包含 searchText
的所有水果.
filterFruits(searchText: "g")
产生 [["Orange", "Orange"]]
如果要搜索所有属性,请使用以下内容:
func filterFruits(searchText: String) ->[[细绳]] {守卫搜索文本!="其他{回果}让针 = searchText.lowercased()返回fruits.filter {fruitObj in返回fruitObj.contains { 属性在attribute.lowercased().contains(needle)}}}
<块引用>
filterFruits(searchText: "g")
产生 [["Apple", "Green"], ["Pear", "Green"], ["Orange", "Orange""]]
为了让您在未来走上正轨:您应该真正引入一个 Fruit
类,该类包含一个特定水果实例的所有相关信息.然后您可以使用第一个函数并执行类似 fruitObj.matches(searchText)
的操作,其中您在 Fruit
类中定义了一个 func
来确定是否水果与搜索匹配.
I have just started to learn swift and i am looking at the tableview and searchbar feature. Below i have my array which is a list of fruits:
var fruits: [[String]] = [["Apple", "Green"],["Pear", "Green"], ["Banana", "Yellow"], ["Orange", "Orange"]]
I have them in a table view with the name of the fruit as the title and the colour as a subtitle. I am trying to use the search bar to filter but i cant seem to get it right. I only want to search for the name of the fruit not the colour.
var filteredFruits = [String]()
var shouldShowSearchResults = false
func searchBar(_ searchBar: UISearchBar, textDidChange searchText: String) {
filteredFruits.removeAll()
var i = 0
while i < fruits.count
{
var filteredFruits = fruits[i].filter ({ (fruit: String) -> Bool in
return fruit.lowercased().range(of: searchText.lowercased()) != nil
})
if searchText != ""
{
shouldShowSearchResults = true
if filteredItems.count > 0
{
filteredFruits.append(filteredItems[0])
filteredItems.removeAll()
}
}
else
{
shouldShowSearchResults = false
}
i += 1
}
self.tableView.reloadData()
}
I do get results returned but it mixes up the subtitles and the titles as well as not returning the correct results. Can anyone point me in the right direction?
I do not understand why you iterate over the fruits using some kind of while loop. Instead I would propose you take advantage of a function like:
func filterFruits(searchText: String) -> [[String]] {
guard searchText != "" else {
return fruits
}
let needle = searchText.lowercased()
return fruits.filter {fruitObj in
return fruitObj.first!.lowercased().contains(needle)
}
}
That function returns all fruits that have a name containing the searchText
.
filterFruits(searchText: "g")
yields[["Orange", "Orange"]]
If you want to search through all attributes use something like:
func filterFruits(searchText: String) -> [[String]] {
guard searchText != "" else {
return fruits
}
let needle = searchText.lowercased()
return fruits.filter {fruitObj in
return fruitObj.contains { attribute in
attribute.lowercased().contains(needle)
}
}
}
filterFruits(searchText: "g")
yields[["Apple", "Green"], ["Pear", "Green"], ["Orange", "Orange"]]
To get you on the right track for the future: you should really introduce a Fruit
class which holds all relevant information of one specific fruit instance. Then you can use the first function and do something like fruitObj.matches(searchText)
where you define a func
inside the Fruit
class which determines if the fruit matches the search.
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