根据OOPer的建议,我将此作为一个单独的问题发布。这是对此的扩展: JSONserialization error
我从一个读取sql数据库的php脚本中提取一个json数组。 json非常大,但应该使用几个在线测试人员进行测试。问题是,当我尝试从数组中取出元素时,它返回nil。它能够下载数据,并且可以正确计算数组中有多少元素,但是当我尝试访问元素时,它返回nil。有什么建议吗?
以下是元素的示例:
{
"id":2045,
"oprettelsesdato":"09-02",
"overskrift":"etc etc etc",
"navn":"xyz xyz xyz",
"tlf":"12345678",
"email":"etc@etc.etc",
"journal":"MJ",
"intro":"yada yada yada yada ",
"annonce":"test",
"address":"testroad"
},
LocationModel.swift
import Foundation
class LocationModel: NSObject {
//properties
var id: String?
var oprettelsesdato: String?
var overskrift: String?
var address: String?
var intro: String?
var email: String?
var tlf: String?
var annonce: String?
var journalnr: String?
override init()
{
}
init(id: String, oprettelsesdato: String, overskrift: String, address: String, intro: String, email: String, tlf: String, annonce: String, journalnr: String) {
self.id = id
self.oprettelsesdato = oprettelsesdato
self.overskrift = overskrift
self.address = address
self.intro = intro
self.email = email
self.tlf = tlf
self.annonce = annonce
self.journalnr = journalnr
}
override var description: String {
return "id: \(id), oprettelsesdato: \(oprettelsesdato), overskrift: \(overskrift), address: \(address), journalnr: \(journalnr)"
}
}
以下是抛出错误的地方:
func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
let item: LocationModel = feedItems[indexPath.row] as! LocationModel
let cellIdentifier: String = "BasicCell"
let myCell: UITableViewCell = tableView.dequeueReusableCell(withIdentifier: cellIdentifier)!
print(feedItems[indexPath.row])
//myCell.detailTextLabel!.text = item.oprettelsesdato
myCell.textLabel!.text = item.overskrift
myCell.textLabel!.leadingAnchor.constraint(equalTo: myCell.leadingAnchor).isActive = true
myCell.textLabel!.topAnchor.constraint(equalTo: myCell.topAnchor).isActive = true
myCell.textLabel!.heightAnchor.constraint(equalTo: myCell.heightAnchor,
multiplier: 1.0).isActive = true
myCell.textLabel!.widthAnchor.constraint(equalTo: myCell.heightAnchor,
multiplier: 0.8).isActive = true
//print(item.id) <-returns nil
//print(item.oprettelsesdato) <-returns nil
//print(item.overskrift) <-returns nil
extralabel!.text = item.oprettelsesdato // <-This is where the error is thrown
}
错误消息:
fatal error: unexpectedly found nil while unwrapping an Optional value
更新 所以我在json解析器中将其缩小到以下内容。尽管所有jsonElements都包含值,但if let optional永远不会成立。怎么了?
for jsonElement in jsonResult {
print(jsonElement["id"]) //<- these print() all return the correct values
print(jsonElement["oprettelsesdato"])
print(jsonElement["overskrift"])
print(jsonElement["address"])
print(jsonElement["intro"])
print(jsonElement["email"])
print(jsonElement["tlf"])
print(jsonElement["annonce"])
print(jsonElement["journal"])
let location = LocationModel()
if let id = jsonElement["id"] as? String,
let oprettelsesdato = jsonElement["oprettelsesdato"] as? String,
let overskrift = jsonElement["overskrift"] as? String,
let address = jsonElement["address"] as? String,
let intro = jsonElement["intro"] as? String,
let email = jsonElement["email"] as? String,
let tlf = jsonElement["tlf"] as? String,
let annonce = jsonElement["annonce"] as? String,
let journalnr = jsonElement["journal"] as? String
{ //this never returns true and location. is never set. Why??
location.id = id
location.oprettelsesdato = oprettelsesdato
location.overskrift = overskrift
location.address = address
location.intro = intro
location.email = email
location.tlf = tlf
location.annonce = annonce
location.journalnr = journalnr
}
locations.append(location)
}
答案 0 :(得分:3)
如果该行
extralabel!.text = item.oprettelsesdato // <-This is where the error is thrown
在打开一个Optional值时,是否会发现&#34;意外发现nil&#34;错误然后原因几乎可以肯定extralabel
(应该命名为extraLabel以遵循camelCase命名约定)为零。在该行设置断点并检查extralabel
。
答案 1 :(得分:2)
您的rowSums
变量是一个可选变量。因为您没有选项说“有值”或“根本没有值”< /strong>。如果您将变量定义为可选,那么要从此变量中获取值,您必须展开它。强力展开的好方法是可选绑定。这是一个例子 - :
var oprettelsesdato:String?
以这种方式尝试您的代码将解决您的问题
只需使用var oprettelsesdato:String?
oprettelsesdato = "hello swift!"
if let value = oprettelsesdato
{
println(value)
}
else
{
println("no value")
}
并在 VARIABLE 中存储阵列值,然后将这些值分配给标签在if let
条件下。
示例 - :强>
if let
答案 2 :(得分:2)
根据给定的JSON id
显然是Int
,而不是String
var id: Int
...
if let id = jsonElement["id"] as? Int,
注意:在提供非可选的初始值设定项时,不要将属性声明为可选属性,也不要在不写入初始值设定项的情况下将其声明为laibiness alibi。 Swift的类型系统鼓励您尽可能使用非可选类型。非可选类型永远不会崩溃应用程序!