如何制作包含 | 的 NSURL （管道字符）？

发布于 2024-09-05 19:47:57 字数 356 浏览 9 评论 0原文

我正在尝试从我的 iPhone 应用程序访问谷歌地图的正向地理编码服务。当我尝试从带有管道的字符串创建 NSURL 时，我只得到一个 nil 指针。

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

我在 google api 中没有看到任何其他方式可以不用管道发送边界坐标。关于我如何做到这一点有什么想法吗？

原文

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+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?

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城歌 2024-09-12 19:47:57

您是否尝试过用 %7C （字符 | 的 URL 编码值）替换管道？

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愁以何悠 2024-09-12 19:47:57

由于 stringByAddingPercentEscapesUsingEncoding 已弃用，您应该使用 stringByAddingPercentEncodingWithAllowedCharacters。

Swift 答案：

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)

编辑：Swift 3 答案：

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)
}

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)
}

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这个俗人 2024-09-12 19:47:57

如果您想保证将来输入的任何奇怪字符的安全，请使用 stringByAddingPercentEscapesUsingEncoding 方法使字符串“URL 友好”...

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];

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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