Ios NSURL返回零

我不知道为什么，但当我调用以下URL时 它给了我错误的请求-无效的URL，尽管这个URL在safari浏览器和其他浏览器上运行良好

http://www.ysl.com/wx/shop-product/women/top-handles#{"ytosQuery":"true","department":"handbags_tophandle_w","gender":"D","brand":"","macro":"","micro":"","season":"A,P,E","color":"","size":"","site":"","section":"","sortRule":"","yurirulename":"searchwithdepartment","microcolor":"","agerange":"","macroMarchio":"","page":"2","productsPerPage":"50","modelnames":"","look":"","washtype":"","fabric":"","prints":"","suggestion":"false","suggestionValue":"","material":"","occasion":"","weight":"","gal

我正在使用以下代码：

NSString *str = [NSString stringWithFormat:@"%@",[payload stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]]];

NSString* webStringURL = [str stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

webStringURL = [str stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

NSURL *payload = [NSURL URLWithString:webStringURL];

---

有人帮忙吗？我做错了什么？

不要在孔URL上使用

stringbyreplacingpercentescapesusingencode:

，而只在GET参数上使用

---

现在

http://

也将被转义，从而成为

http%3A%2F%2F

，它作为URL无效。

{

在URL中是不合法的。事实上，这是JSON吗？如果你看到了，我也在整个URL中添加了PercenteScape。你不应该，在参数上应该转义。你根本不应该使用这两种方法中的任何一种。第一种方法删除转义序列（但没有转义序列），第二个尝试将它们添加到所有位置。正确的URL数据编码方法在苹果开发者网站上有介绍。在返回ARC的过程中，您还需要（u bridge CFStringRef）您的NSString和（u bridge_transfer NSString*）它们。