Question

我正在创建segue，以便将数据从tableview控制器传递到详细视图控制器as shown in the picture。我的表格单元格有一个UIImage和两个UILabel，我想在详细视图控制器中显示其中三个。但是，当我运行该程序时，它显示为this。

这是tableviewcontroller.m

的segue

-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
    NSLog(@"prepareForSegue:%@", segue.identifier);



    if([segue.identifier isEqualToString:@"TeamMembersSegue"]){

        NSIndexPath*indexPath = (NSIndexPath*)sender;

        ViewController2*vc = ((UINavigationController*)segue.destinationViewController).topViewController;

        vc.lawfirmname = [tableData objectAtIndex:indexPath.row];
        vc.detail= [detail objectAtIndex:indexPath.row];
        vc.lawyerimage = [thumbnails objectAtIndex:indexPath.row];
}
}

和detailviewcontroller.h的代码（视图控制器2）

#import <UIKit/UIKit.h>


@interface ViewController2 : UIViewController

@property (nonatomic, strong) NSString *lawfirmname;
@property (strong, nonatomic) IBOutlet UILabel *lawfirm;

@property (nonatomic, strong) NSString *detail;
@property (weak, nonatomic) IBOutlet UITextView *detailtextview;

@property (nonatomic, strong) UIImage*lawyerimage;
@property (weak, nonatomic) IBOutlet UIImageView *image;

@end

和detailviewcontroller.m（查看控制器2）

#import "ViewController2.h"

    @interface ViewController2 ()

    @end

    @implementation ViewController2


    @synthesize lawfirmname;
    @synthesize lawfirm;
    @synthesize detail;
    @synthesize detailtextview;
    @synthesize lawyerimage;
    @synthesize image;

    - (void)viewDidLoad {
        [super viewDidLoad];


        // Do any additional setup after loading the view.

        self.lawfirm.text = self.lawfirmname;
        self.detailtextview.text = self.detail;
        self.image.image = self.lawyerimage;
    }

    - (void)didReceiveMemoryWarning {
        [super didReceiveMemoryWarning];
        // Dispose of any resources that can be recreated.
    }



    /*
    #pragma mark - Navigation

    // In a storyboard-based application, you will often want to do a little preparation before navigation
    - (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
        // Get the new view controller using [segue destinationViewController].
        // Pass the selected object to the new view controller.
    }
    */

    @end

如果有人有任何解决方案，我将不胜感激。如果您不理解这个问题，请通知我。

Answer 1

使用 didSelectRowAtIndexPath 将数据从tableview控制器传递到详细信息视图控制器。例如： -

- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath
{
    self.selectedRowModel = [self.someArray objectAtIndex:indexPath.row];
    [self performSegueWithIdentifier:@"TeamMembersSegue" sender:self];
}

然后覆盖 prepareForSegue 方法

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if([segue.identifier isEqualToString:@"TeamMembersSegue"]){
        ViewController2 *vc = segue.destinationViewController;
        vc.selectedModel = self.selectedRowModel;
    }
}

注意： - 创建一个公共行模型来表示您的各个Row项。 Model将为UITableView中的每一行存储图像URL和图像描述。

Answer 2

现在，在我将segue改为

之后

    - (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath{
        [self performSegueWithIdentifier:@"showlawfirmdetail" sender:indexPath];
    }

-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
    NSLog(@"prepareForSegue:%@", segue.identifier);



    if([segue.identifier isEqualToString:@"showlawfirmdetail"]){

        NSIndexPath*indexPath = (NSIndexPath*)sender;

        ViewController2*vc = (ViewController2*)segue.destinationViewController;

        vc.lawfirmname = [tableData objectAtIndex:indexPath.row];
        vc.detail= [detail objectAtIndex:indexPath.row];
        vc.lawyerimage = [thumbnails objectAtIndex:indexPath.row];
}
}

部分工作。剩下的问题是我无法在视图控制器2中添加图像。它崩溃了。但是，如果我删除代码self.image.image = self.lawyerimage;，则应用程序可以顺利运行。

Answer 3

@Herman又一枪。我创建了一个项目，并重复了所有代码，只是为了看看我是否得到了我不是的东西。所以我没弄清楚它出了什么问题。

我认为有一件事是可能的，如果你由于某种原因丢失了IBOutlet连接或者它是重复的。检查出来，就像在图像中一样。在右上角看到只有一个推荐出口：

这里是我的“TableViewController.h”代码，我做的一件事就是以编程方式设置NSStrings和UIImage：

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {

    if([segue.identifier isEqualToString:@"TeamMembersSegue"]){
        ViewController2 *vc = segue.destinationViewController;
        vc.lawfirmname = @"Title";
        vc.detail = @"Detail here";
        vc.lawyerimage = [UIImage imageNamed:@"print"];
    }

}

我的“ViewController2.h”和你的一样：

- (void)viewDidLoad {
    [super viewDidLoad];
    // Do any additional setup after loading the view.

    self.lawfirm.text = self.lawfirmname;
    self.detailtextview.text = self.detail;
    self.image.image = self.lawyerimage;
}

看，它工作正常：

Answer 4

我有解决方案：替换

vc.lawyerimage = [thumbnails objectAtIndex:indexPath.row];

通过

    vc.lawyerimage = [UIImage imageNamed:[thumbnails objectAtIndex:indexPath.row]];

Answer 5

我建议你在第二个视图控制器中声明三个不同的字符串，然后通过prepareForSegue方法将单元格的数据传递给这些字符串。在此之后，使用IBOutlets或viewDidLoad方法填写这三个字符串中的viewWillAppear。这解决了我的问题。希望这有用。

将数据从tableview控制器传递到视图控制器，但它显示为null

5 个答案: