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How to make an NSURL that contains a | (pipe character)?

开发者 https://www.devze.com 2023-01-03 05:38 出处:网络
I am trying to access google maps\' forward geocoding service from my iphone app. When i try to make an NSURL from a string with a pipe in it I just get a nil pointer.

I am trying to access google maps' forward geocoding service from my iphone app. When i try to make an NSURL from a string with a pipe in it I just get a nil pointer.

NSURL *searchURL = [NSURL URLWithString:@"http://maps.google.com/maps/api/geocode/json?address=6th+开发者_Go百科and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false"];

I dont see any other way in the google api to send bounds coordinates with out a pipe. Any ideas about how I can do this?


Have you tried replacing the pipe with %7C (the URL encoded value for the char |)?


As stringByAddingPercentEscapesUsingEncoding is deprecated, you should use stringByAddingPercentEncodingWithAllowedCharacters.

Swift answer:

let rawUrlStr = "http://maps.google.com/maps/api/geocode/json?address=6th+and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false";
let urlEncoded = rawUrlStr.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet())
let url = NSURL(string: urlEncoded)

Edit: Swift 3 answer:

let rawUrlStr = "http://maps.google.com/maps/api/geocode/json?address=6th+and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false";
if let urlEncoded = rawUrlStr.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) {
    let url = NSURL(string: urlEncoded)
}


If you want to be safe for whatever weird characters you will put in the future, use stringByAddingPercentEscapesUsingEncoding method to make the string "URL-Friendly"...

NSString *rawUrlStr = @"http://maps.google.com/maps/api/geocode/json?address=6th+and+pine&bounds=37.331689,-122.030731|37.331689,-122.030731&sensor=false";
NSString *urlStr = [rawUrlStr stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *searchURL = [NSURL URLWithString:urlStr];
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