iOS：WebView加载网址 [英] iOS: WebView Loading a url

问题描述

我正在尝试在 UIWebView 中打开以下网址，但无法加载，而是将其更改为：

I am trying to open the following url in UIWebView but it fails to load whereas changing it to:

 http://www.google.com

工作正常。

我要加载的网址是：

[webView loadRequest:[NSURLRequest requestWithURL:[NSURL URLWithString:[NSString stringWithFormat:@"%@%@%@%@%@",@"http://m.forrent.com/search.php?address=",[[bookListing objectForKey:@"Data"] objectForKey:@"zip"],@"&beds=&baths=&price_to=0#{\"lat\":\"0\",\"lon\":\"0\",\"distance\":\"25\",\"seed\":\"1622727896\",\"is_sort_default\":\"1\",\"sort_by\":\"\",\"page\":\"1\",\"startIndex\":\"0\",\"address\":\"",[[bookListing objectForKey:@"Data"] objectForKey:@"zip"],@"\",\"beds\":\"\",\"baths\":\"\",\"price_to\":\"0\"}"]]]];

更新：

我故意逃脱双引号，否则它给我一个错误。
我通过在浏览器中打开（在笔记本电脑上）检查了网址，它完全正常：

I have purposely escaped the double quotes otherwise it gives me an error. I checked the url by opening in my browser (on laptop) and it works perfectly fine:

浏览器中的网址：

http://m.forrent.com/search.php?address=92115&beds=&baths=&price_to=0#{%22lat%22:%220%22,%22lon%22:%220%22,%22distance%22:%2225%22,%22seed%22:%221622727896%22,%22is_sort_default%22:%221%22,%22sort_by%22:%22%22,%22page%22:%221%22,%22startIndex%22:%220%22,%22address%22:%2292115%22,%22beds%22:%22%22,%22baths%22:%22%22,%22price_to%22:%220%22}

推荐答案

您的代码行看起来很复杂，但基本上它是一个非常简单的代码。

Your line of code looks convoluted, but basically it is a very simple one.

您应该将此代码从一行分解为多行更具可读性。
这也将允许您记录并检查您实际创建的URL，如下所示：

You should breakup this code from a one liner to multiple lines that are more readable. That will also allow you to log and check the URL you actually created, like so:

NSLog（@我的网址： ％@，urlString）;

更新：
我看到你添加了完整的网址。 Webview确实无法加载该URL（UIWebkit错误101）。

Update: I see you added the full url. Webview indeed fails to load that url (UIWebkit error 101).

导致问题的url部分是params后面的'＃'字符和字典。你应该url编码网址的那一部分。

The part of the url that causes the problem is the '#' character and dictionary that follows in the params. You should url encode that part of the url.

试试这个：

NSString *address = @"http://m.forrent.com/search.php?";
NSString *params1 = @"address=92115&beds=&baths=&price_to=0";

// URL encode the problematic part of the url.
NSString *params2 = @"#{%22lat%22:%220%22,%22lon%22:%220%22,%22distance%22:%2225%22,%22seed%22:%221622727896%22,%22is_sort_default%22:%221%22,%22sort_by%22:%22%22,%22page%22:%221%22,%22startIndex%22:%220%22,%22address%22:%2292115%22,%22beds%22:%22%22,%22baths%22:%22%22,%22price_to%22:%220%22}";
params2 = [self escape:params2];

// Build the url and loadRequest
NSString *urlString = [NSString stringWithFormat:@"%@%@%@",address,params1,params2];
[self.webView loadRequest:[NSURLRequest requestWithURL:[NSURL URLWithString:urlString]]];

我使用的转义方法：

- (NSString *)escape:(NSString *)text
{
    return (__bridge NSString *)CFURLCreateStringByAddingPercentEscapes(NULL,
                                                                        (__bridge CFStringRef)text, NULL,
                                                                        (CFStringRef)@"!*'();:@&=+$,/?%#[]",
                                                                        kCFStringEncodingUTF8);
}

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