文本不显示 Var. Swift 的值

Question

我是初学者，但我试图在文本字段中显示"valueWeWantToGrab"，但没有显示结果。

这是我显示某人的Instagram关注者数量的代码

struct ContentView: View {
@State var valueWeWantToGrab = ""
var body: some View {

let baseUrl = "http://www.instagram.com/"
let username = ""
let url = URL(string: baseUrl + username)!
let task = URLSession.shared.dataTask(with: url) { (data, response, error) in
guard let data = data else {
print("data was nil")
return
}
guard let htmlString = String(data: data, encoding: .utf8) else {
print("couldn't cast data into String")
return
}
print(htmlString)
let leftSideString = """
edge_followed_by":{"count":
"""
let rightSideString = """
},"followed_by_viewer
"""
guard
let leftSideRange = htmlString.range(of: leftSideString)
else {
print("couldn't find left range")
return
}
guard
let rightSideRange = htmlString.range(of: rightSideString)
else {
print("couldn't find right range")
return
}
let rangeOfTheData = leftSideRange.upperBound..<rightSideRange.lowerBound
var valueWeWantToGrab = htmlString[rangeOfTheData]
print(valueWeWantToGrab) // prints the follower count: 19093
}
Text(valueWeWantToGrab)
}

}

Answer 1

问题是你声明变量valueWeWantToGrab两次：一次在顶部用@State注释，一次在body变量的getter函数中朝向底部。 您实际上想在底部做的是分配给在顶部声明的变量，而不是通过从行var valueWeWantToGrab = htmlString[rangeOfTheData]中删除var来声明新变量。这样做，您还必须使用字符串初始值设定项(String())包装htmlString[rangeOfTheData]，因为htmlString[rangeOfTheData]为您提供Substring类型的值，而不是String类型的值。结果如下：

...
let rangeOfTheData = leftSideRange.upperBound..<rightSideRange.lowerBound
valueWeWantToGrab = String(htmlString[rangeOfTheData])
print(valueWeWantToGrab) // prints the follower count: 19093
...

此外，为了使您的代码更干净(尽管不是使其工作所必需的)，最好将加载数据的代码移动到专用方法中，而不是将其放在body变量的 getter 中(作为一般经验法则，尝试只将 UI 元素放在 SwiftUIViews 的body中，并在其他地方执行业务逻辑 - 这将帮助您保持代码干净易读)。然后，您可以在ContentView的初始值设定项中调用该方法。这样做的好处是，您可以在其他地方重用该方法 - 例如，将来实现刷新按钮。最终结果可能如下所示：

struct ContentView: View {
@State var valueWeWantToGrab = ""

var body: some View {
Text(valueWeWantToGrab)
}

init() {
loadFollowerCount()
}

private func loadFollowerCount() {
let baseUrl = "http://www.instagram.com/"
let username = ""
let url = URL(string: baseUrl + username)!
URLSession.shared.dataTask(with: url) { (data, response, error) in
guard let data = data else {
print("data was nil")
return
}
guard let htmlString = String(data: data, encoding: .utf8) else {
print("couldn't cast data into String")
return
}
print(htmlString)
let leftSideString = """
edge_followed_by":{"count":
"""
let rightSideString = """
},"followed_by_viewer
"""
guard
let leftSideRange = htmlString.range(of: leftSideString)
else {
print("couldn't find left range")
return
}
guard
let rightSideRange = htmlString.range(of: rightSideString)
else {
print("couldn't find right range")
return
}
let rangeOfTheData = leftSideRange.upperBound..<rightSideRange.lowerBound
valueWeWantToGrab = String(htmlString[rangeOfTheData])
print(valueWeWantToGrab) // prints the follower count: 19093
}
}
}

如果你来自 Web 开发背景(这就是我的猜测)，最后一个提示：Swift 中的var与 JavaScript 中的var不同。在 JavaScript 中，重新声明将覆盖以前的valueWeWantToGrab值，并且您的代码将起作用。然而，在 Swift 中，这实际上在 getter 函数的局部范围内声明了一个新的、不同的变量。查看 Swift 文档，了解有关变量声明在 Swift 中如何工作的更多信息。