Question

我正在学习swift，我需要发出GET请求，所以这是我的代码：

    var j = ["user": "xxxxxx@xxxxxx.xxx"]
    var e: NSError?
    let jsonData = NSJSONSerialization.dataWithJSONObject(j,options: NSJSONWritingOptions(0),error: &e) 
    var jsonString = NSString(data: jsonData!, encoding: NSUTF8StringEncoding)

    var urlPath = "http://test.example.io/materials?query=\(jsonString)";
    println(urlPath)

    var url: NSURL = NSURL(string: urlPath)
    var request: NSMutableURLRequest =  NSMutableURLRequest(URL: url)
    request.addValue("xxxxxxxxxxxxxxxxxxxx" , forHTTPHeaderField: "Teech-Application-Id")
    request.addValue("xxxxxxxxxxxxxxxxxxxx" , forHTTPHeaderField: "Teech-REST-API-Key")
    request.addValue("application/json" , forHTTPHeaderField: "Content-Type")
    request.HTTPMethod = "GET"

    println(request)

这是我的输出调试

http://test.example.io/materials?query={"user":"xxxxxxxxx@xxxxxxxxx.xxxxx"}

<NSMutableURLRequest: 0x7f8ee3480320> { URL: (null), headers: {
"Content-Type" = "application/json";
"Teech-Application-Id" = xxxxxxxxxxxxxxxxxxxxxxxx;
"Teech-REST-API-Key" = xxxxxxxxxxxxxxxxxxxxxxxxx;
} }

为什么第一个打印函数返回正确的url并且在打印请求中有一个空url，哪里是我的错误？我已经尝试过没有json的url并且它运行了。

Answer 1

我完成了，解决方案是逃避字符串。

var escapedSearchTerm = urlPath.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
var url: NSURL = NSURL(string: escapedSearchTerm!)

获取与Json ios的URL连接 - swift

1 个答案: