ios - 获取键的方法 JSON 响应(字符串)编码投诉
<p><p>我的ResponseString如下,</p>
<pre><code>SUCCESS:
{"code":200,"shop_detail":{"name":"dad","address":"556666"},
"shop_types : [{"name":"IT\/SOFTWARE","merchant_type":"office"}]}
</code></pre>
<p>我的带有headers的Get请求代码如下,</p>
<pre><code>func getProfileAPI() {
let headers: HTTPHeaders = [
"Authorisation": AuthService.instance.tokenId ?? "",
"Content-Type": "application/json",
"Accept": "application/json"
]
print(headers)
let scriptUrl = "http://haitch.igenuz.com/api/merchant/profile"
if let url = URL(string: scriptUrl) {
var urlRequest = URLRequest(url: url)
urlRequest.httpMethod = HTTPMethod.get.rawValue
urlRequest.addValue(AuthService.instance.tokenId ?? "", forHTTPHeaderField: "Authorization")
urlRequest.addValue("application/json", forHTTPHeaderField: "Content-Type")
urlRequest.addValue("application/json", forHTTPHeaderField: "Accept")
Alamofire.request(urlRequest)
.responseString { response in
debugPrint(response)
print(response)
if let result = response.result.value // getting the json value from the server
{
print(result)
let jsonData1 = result as NSString
print(jsonData1)
let name = jsonData1.object(forKey: "code") as!
print(name)
// var data = jsonData1!["shop_detail"]?["name"] as? String
} }
}
</code></pre>
<p>当我尝试获取“name”的值时,它得到'[<__NSCFString 0x7b40f400> valueForUndefinedKey:]:此类对于键代码不符合键值编码。请指导我获取名称,地址的值..?????</p></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>您可以使用<strong>Response Handler</strong> 代替<strong>Response String Handler</strong>:</p>
<blockquote>
<p><strong>Response Handler</strong></p>
<p>The response handler does NOT evaluate any of the response data. It
merely forwards on all information directly from the URL session
delegate. It is the Alamofire equivalent of using cURL to execute a Request.</p>
</blockquote>
<pre><code>struct Root: Codable {
let code: Int
let shopDetail: ShopDetail
let shopTypes:
}
struct ShopDetail: Codable {
let name, address: String
}
struct ShopType: Codable {
let name, merchantType: String
}
</code></pre>
<p>如果您将解码器 <code>keyDecodingStrategy</code> (检查 <a href="https://www.hackingwithswift.com/articles/52/swift-4-1-improves-codable-with-keydecodingstrategy" rel="noreferrer noopener nofollow">this</a> )设置为 <code>.convertFromSnakeCase</code> ,您也可以省略结构声明中的编码键,正如@评论中已经提到的那样瓦迪安:</p>
<hr/>
<pre><code>Alamofire.request(urlRequest).response { response in
guard
let data = response.data,
let json = String(data: data, encoding: .utf8)
else { return }
print("json:", json)
do {
let decoder = JSONDecoder()
decoder.keyDecodingStrategy = .convertFromSnakeCase
let root = try decoder.decode(Root.self, from: data)
print(root.shopDetail.name)
print(root.shopDetail.address)
for shop in root.shopTypes {
print(shop.name)
print(shop.merchantType)
}
} catch {
print(error)
}
}
</code></pre>
<p>有关编码和解码自定义类型的更多信息,您可以阅读此 <a href="https://benscheirman.com/2017/06/swift-json/" rel="noreferrer noopener nofollow">post</a> .</p></p>
<p style="font-size: 20px;">关于ios - 获取键的方法 JSON 响应(字符串)编码投诉,我们在Stack Overflow上找到一个类似的问题:
<a href="https://stackoverflow.com/questions/53459301/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/53459301/
</a>
</p>
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