Question

我有2个视图控制器，我将数据从1传递到2但是2中的NSString为空（null）我做错了什么？

查看控制器1 .m

    - (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView
{
    NSLog(@"Method 1");
    return 1;
}

- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section
{
    NSLog(@"Method 2");
    NSLog(@"%lu", (unsigned long)[allPosts count]);
    return [[[allPosts objectForKey:@"posts"] valueForKey:@"title"] count];
}

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
    NSLog(@"Method 3");
    static NSString *CellIdentifier = @"Cell";

    UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier];

    if (cell == nil) {
        cell = [[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier];
    }

    cell.textLabel.text = [NSString stringWithFormat:@"%@", [[[allPosts objectForKey:@"posts"] valueForKey:@"title"] objectAtIndex:indexPath.row]];

    return cell;
}

- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath
{
    OldPostViewController *vc = [[OldPostViewController alloc] init];
    NSString *url = [NSString stringWithFormat:@"server/app/post.php?id=%@", [[[allPosts objectForKey:@"posts"] valueForKey:@"id"] objectAtIndex:indexPath.row]];
    vc.postURL = url;
    NSLog(@"%@", vc.postURL);
    [self performSegueWithIdentifier:@"viewPost" sender:self];
}

查看控制器2 .h

    #import <UIKit/UIKit.h>

@interface OldPostViewController : UIViewController <UIWebViewDelegate>
@property (weak, nonatomic) IBOutlet UIWebView *blogView;
@property (strong, nonatomic) NSString *postURL;

@end

查看控制器2 .m

NSLog(@"URL: %@", _postURL);

NSString *fullURL = _postURL;
NSURL *url = [NSURL URLWithString:fullURL];
NSURLRequest *requestObj = [NSURLRequest requestWithURL:url];
[blogView loadRequest:requestObj];

任何人都可以告诉我我可以做什么，因为我已经尝试了我可以在Google和Google上找到的所有解决方案。计算器

Answer 1

你不能这样做。当您致电performSegueWithIdentifier:时，这将不会使用您在本地初始化的vc。

实施prepareForSegue:来执行此操作。像：

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([[segue identifier] isEqualToString:@"viewPost"])
    {
        OldPostViewController *vc= (OldPostViewController *)[segue destinationViewController];
        vc.postURL = yourURL;
    }
}

或

你可以这样做：

- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { OldPostViewController *vc = [[OldPostViewController alloc] init]; NSString *url = [NSString stringWithFormat:@"server/app/post.php?id=%@", [[[allPosts objectForKey:@"posts"] valueForKey:@"id"] objectAtIndex:indexPath.row]]; vc.postURL = url; NSLog(@"%@", vc.postURL); [self presentViewController:vc animated:YES completion:nil]; }

Answer 2

您需要将didSelect方法中的URL作为属性，然后在prepareSegue方法中访问它

抱歉，我现在无法编写完整的代码：）

从视图控制器传递的数据返回null

2 个答案: