查看JSON feed Swift 3.0

我想在IOS应用程序中查看我的php结果。代码如下：查看JSON feed Swift 3.0

func get() 
    { 
     //let url = NSURL(string: "http://www.w3schools.com/js/customers_mysql.php") 
     let url = NSURL(string: "http://www.helpmewastetime.com/service.php") 
     let data = NSData(contentsOf: url! as URL) 

     values = try! JSONSerialization.jsonObject(with: data! as Data, options: JSONSerialization.ReadingOptions.allowFragments) as! NSArray 


     var array = [""] 
     array = (values.value(forKey: "Name") as? [String])! 
     lb_Value.text = array[0] 


    }

的代码为W3Schools的链接工作正常，但是它崩溃时，我用我的链接（在去爸爸托管）如果您打开一个网页浏览器的链接，我们得到预期的结果。

这是收到的错误：

2016-11-30 16:44:24.977781 Fun iOS App[2590:112849] [] __nw_connection_get_connected_socket_block_invoke 2 Connection has no connected handler 
2016-11-30 16:44:24.979616 Fun iOS App[2590:112846] PAC stream failed with 
2016-11-30 16:44:24.981519 Fun iOS App[2590:112849] [] nw_proxy_resolver_create_parsed_array PAC evaluation error: kCFErrorDomainCFNetwork: 2 
2016-11-30 16:44:26.145018 Fun iOS App[2590:112849] [] nw_endpoint_handler_add_write_request [1.1.1 107.180.54.250:80 failed socket-flow (satisfied)] cannot accept write requests 
2016-11-30 16:44:26.145764 Fun iOS App[2590:112847] [] __tcp_connection_write_eof_block_invoke Write close callback received error: [22] Invalid argument 
2016-11-30 16:44:26.150620 Fun iOS App[2590:112846] [] nw_proxy_resolver_create_parsed_array PAC evaluation error: kCFErrorDomainCFNetwork: 2 
fatal error: 'try!' expression unexpectedly raised an error: Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}: file /Library/Caches/com.apple.xbs/Sources/swiftlang/swiftlang-800.0.58.6/src/swift/stdlib/public/core/ErrorType.swift, line 178

任何帮助，将不胜感激

来源

2016-11-30 Nassif Bousaba

http://www.helpmewastetime.com/service.php指向一个[HTML页面]（https://www.evernote.com/l/AOyy9a1kG7xDtbawSnrbB41nOVEqpXWDYnc），而不是一个JSON响应... – Moritz

正如埃里克·绫在他的评论中提到。问题是您的PHP服务将JSON数据吐出到HTML页面中。

当我查看网页源我得到如下：

<html> 
[{"ID":"1","Name":"Peter"},{"ID":"2","Name":"James"}] 
</html>

W3Schools的URL的网页源代码仅仅是JSON。在你的PHP服务中，用JSON输出你的数据，你应该很好。

来源

2016-11-30 15:14:57 toddg

谢谢托德！现在工作正常 –

查看JSON feed Swift 3.0

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