Question

我正在开发一个新闻应用程序，从rss提要中获取内容。当我点击TableViewCell时，我将带有标题，链接等的NSDictionary对象传递给下一个ViewController。在ViewController中我定义了NSDictionary * item;我可以通过设置viewcontroller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，当我分配像这个@"http://somesite.com"这样的地址时，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。正如我所提到的，我检查了[self.item objectForKey:@"link"];的值是一个有效的URL，因为它在导航栏的标题中显示它。

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。

我做错了什么？

Answer 1

您在哪里为self.title分配[item objectForKey:@"link"]？

您是否在.h文件中添加了item作为视图控制器的属性？如果您还没有使用上面发布的viewDidLoad方法，则无法访问self.item。

在你的.h文件中：

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

在你的.m文件中（调整它以适合你的init方法）：

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

执行此操作后，您将通过item在视图控制器的方法中对self.item属性进行全局访问。

Answer 2

它已经解决了，我做了

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

现在有效，显然我在网址之后发送了一些空格，这就是为什么它不是一个有效的网址，即使它有效。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

无法在UIWebView中加载页面

2 个答案: