iOS HTTP请求 URL中含有特殊字符 !$&'()*+,;= 处理
有些符号在URL中是不能直接传递的，如果要在URL中传递这些特殊符号，那么就要使用他们的编码了。
解决办法：
对参数做编码，代码如下：
-(NSString*)encodeUrlParam
{
static NSString * const kAFCharactersGeneralDelimitersToEncode = @":#[]@"; // does not include "?" or "/" due to RFC 3986 - Section 3.4
staticNSString*constkAFCharactersSubDelimitersToEncode =@"!$&'()*+,;=";
NSMutableCharacterSet * allowedCharacterSet = [[NSCharacterSet URLQueryAllowedCharacterSet] mutableCopy];
[allowedCharacterSetremoveCharactersInString:[kAFCharactersGeneralDelimitersToEncodestringByAppendingString:kAFCharactersSubDelimitersToEncode]];
// FIXME: https://github.com/AFNetworking/AFNetworking/pull/3028
// return [string stringByAddingPercentEncodingWithAllowedCharacters:allowedCharacterSet];
staticNSUIntegerconstbatchSize =50;
NSUIntegerindex =0;
NSMutableString*escaped =@"".mutableCopy;
while(index
#pragma GCC diagnostic push
#pragma GCC diagnostic ignored"-Wgnu"
NSUIntegerlength =MIN(self.length - index, batchSize);
#pragma GCC diagnostic pop
NSRangerange =NSMakeRange(index, length);
range = [self rangeOfComposedCharacterSequencesForRange:range];
NSString*substring = [selfsubstringWithRange:range];
NSString*encoded = [substringstringByAddingPercentEncodingWithAllowedCharacters:allowedCharacterSet];
[escapedappendString:encoded];
index += range.length;
}
returnescaped;
}
