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我正在开发一个从 rss 提要中获取内容的新闻应用程序。当我单击 TableViewCell 时，我将带有标题、链接等的 NSDictionary 对象传递给下一个 ViewController。在 ViewController 我定义 NSDictionary *item; 我可以通过像这样设置视图控制器的标题来验证值是否正确传递： self.title = [item objectForKey:@"link"]; 标题显示了我试图在我的 UIWebView 中打开的链接。

这是我的 ViewController 的实现

    //
//  NewsDetailViewController.m
//  NewsApp2
//
//  Created by Marco Soria on 12/27/10.
//  Copyright 2010 __MyCompanyName__. All rights reserved.
//

#import "NewsDetailViewController.h"


@implementation NewsDetailViewController

@synthesize item, itemTitle, itemDate, itemSummary;

-(id)initWithItem:(NSDictionary *)theItem{
    if(self = [super initWithNibName:@"NewsDetail" bundle:[NSBundle mainBundle]]){
        self.item = theItem;
        self.title = [item objectForKey:@"link"];
    }

    return self;
}

// The designated initializer.  Override if you create the controller programmatically and want to perform customization that is not appropriate for viewDidLoad.
/*
- (id)initWithNibName:(NSString *)nibNameOrNil bundle:(NSBundle *)nibBundleOrNil {
    self = [super initWithNibName:nibNameOrNil bundle:nibBundleOrNil];
    if (self) {
        // Custom initialization.
    }
    return self;
}
*/


// Implement viewDidLoad to do additional setup after loading the view, typically from a nib.
- (void)viewDidLoad {
    [super viewDidLoad];

    NSString *urlAddress = [self.item objectForKey:@"link"]; 

    NSURL *baseURL = [[NSURL URLWithString: urlAddress] retain];        

    NSURLRequest *request = [NSURLRequest requestWithURL:baseURL];

    [self.itemSummary loadHTMLString:urlAddress baseURL:nil];
    [self.itemSummary loadRequest:request];

    [baseURL release];
}


/*
// Override to allow orientations other than the default portrait orientation.
- (BOOL)shouldAutorotateToInterfaceOrientation:(UIInterfaceOrientation)interfaceOrientation {
    // Return YES for supported orientations.
    return (interfaceOrientation == UIInterfaceOrientationPortrait);
}
*/

- (void)didReceiveMemoryWarning {
    // Releases the view if it doesn't have a superview.
    [super didReceiveMemoryWarning];

    // Release any cached data, images, etc. that aren't in use.
}

- (void)viewDidUnload {
    [super viewDidUnload];
    // Release any retained subviews of the main view.
    // e.g. self.myOutlet = nil;
}


- (void)dealloc {
    [super dealloc];

}


@end

现在，当我像这样分配地址时@"http://somesite.com"，UIWebView 加载得很好，但是当我这样做时：NSString *urlAddress = [self.item objectForKey:@"link"]; 它永远不会加载。正如我提到的，我检查了 的值[self.item objectForKey:@"link"];是一个有效的 url，因为它显示在导航栏的标题中。

如果我这样做： [self.itemSummary loadHTMLString:urlAddress baseURL:nil]; UIWebView 显示 url，另一种验证 urlAddress 是否具有正确 url 的方法。

我究竟做错了什么？

4

2 回答 2

0

self.title你在哪里分配[item objectForKey:@"link"]？

您是否item在 .h 文件中添加了视图控制器的属性？如果您没有在viewDidLoad上面发布的方法中使用，您将无法访问self.item.

在您的 .h 文件中：

@interface MyViewController : UIViewController {
    NSDictionary *item;
}
@property (nonatomic, readwrite, retain) NSDictionary *item;

在您的 .m 文件中（调整它以适合您的 init 方法）：

-(id)initWithItem:(NSDictionary *)item {
    if (self = [super initWithNibName: nil bundle: nil]) {
        self.item = item;   
    }
    return self;
}

完成此操作后，您将在视图控制器的方法中item通过self.item.

于 2010-12-29T21:32:25.150 回答
0

解决了，我解决了

NSURL *baseURL = [[NSURL URLWithString: [urlAddress stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]] retain];

现在可以工作了，显然我在 url 之后发送了一些空格，这就是为什么它不是有效的 url，即使它看起来有效。我 NSLog 并得到了 baseUrl http://site.com/item.php?id=2458%20%0A%09%20%20%20%20%20%20%20%20%09%20%20%20 %20%09%09%09%09

于 2010-12-30T04:01:56.200 回答