我正在使用IOS 5和Storyboard编码。我已经构建了一个应用程序,其中我有一个tableView连接到带有搜索栏的sqlite数据库。 当我们触摸一行时,它会自动转到另一个名为“详细信息”的视图控制器。 我需要将数据从我的表视图传递到详细信息视图控制器,例如将author.title传递给labelText.text字段。 有什么想法吗?
编辑问题:
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{
// NSString *Title;
Details *dv = (Details*)segue.destinationViewController;
author.title = dv.labelText.text;
}
部分代码:
//
// Details.m
// AuthorsApp
//
// Created by georges ouyoun on 7/17/12.
// Copyright (c) 2012 __MyCompanyName__. All rights reserved.
//
#import "Details.h"
#import "Author.h"
@interface Details ()
@end
@implementation Details
@synthesize labelText;
@synthesize selectedAuthors;
@synthesize author;
- (id)initWithNibName:(NSString *)nibNameOrNil bundle:(NSBundle *)nibBundleOrNil
{
self = [super initWithNibName:nibNameOrNil bundle:nibBundleOrNil];
if (self) {
// Custom initialization
}
return self;
}
- (void)viewDidLoad
{
[super viewDidLoad];
self.labelText.text = self.author.title;
// Do any additional setup after loading the view.
NSLog(@"Everything is ok now !");
}
- (void)viewDidUnload
{
// [self setLabelText:nil];
NSLog(@"U have entered view did unload");
[super viewDidUnload];
[self setLabelText:Nil];
// Release any retained subviews of the main view.
}
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{
// NSString *Title;
Details *dv = (Details*)segue.destinationViewController;
// author.title = dv.labelText.text;
dv.labelText.text = author.title;
}
/*
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{
if ([segue.identifier isEqualToString:@"AuthorsCell"]) {
[segue.destinationViewController setLabelText:author.title];
}
}
/*
-(void)viewWillAppear:(BOOL)animated
{
self.labelText.text = author.title;
NSLog(@"U have entered the viewWillAppear tag");
// detailsLabel.text = food.description;
}
*/
- (BOOL)shouldAutorotateToInterfaceOrientation:(UIInterfaceOrientation)interfaceOrientation
{
return YES;
}
- (void)dealloc {
[labelText release];
[super dealloc];
}
@end
答案 0 :(得分:6)
出于演示目的,我假设我们有两个视图控制器ListViewController
和DetailViewController
。 ListViewController具有tableview,您已创建segue between your tableview cell and details view controller
。
我假设你将它命名为'DetailSegue'。您的故事板应该与下图相似。
选择tableViewCell和DetailViewController之间的segue,打开属性检查器并将标识符指定为“DetailSegue”。见下图,
现在运行应用程序,您应该看到从ListViewController到DetailViewController的平滑导航。
Assuming you datasource is an array of stings and you have your datasource and delegate setup already properly for your list tableView
。
在List视图控制器中添加方法prepareForSegue:sender,只要segue即将执行,就会调用此方法。此方法将自动调用。
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
if ([segue.identifier isEqualToString:@"DetailSegue"]) {
DetailViewController *detailVC = (DetailViewController*)segue.destinationViewController;
NSIndexPath *indexPath = [self.tableView indexPathForCell:(UITableViewCell *)sender];
detailVC.title = [self.source objectAtIndex:indexPath.row];
}
}
在DetailViewController中添加一个名为title
的属性来存储所选标题,并为UILabel添加另一个插座以在视图中显示标题。
DetailViewController.h
#import <Foundation/Foundation.h>
@interface RecipeListViewController : NSObject
@property (weak, nonatomic) NSString *title;
@property (weak, nonatomic) IBOutlet UILabel *titleLabel;
@end
DetailViewController.m
@implementation RecipeListViewController
- (void)viewDidLoad
{
[super viewDidLoad];
self.titleLabel.text = self.title;
}
@end
不要忘记连接插座。我希望这能让您大致了解如何在视图控制器之间传递数据。
答案 1 :(得分:0)
是的,您需要使用
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
方法(覆盖)女巫允许你通过获取segue.destinationViewController来传递细节,然后设置目标控制器的属性。
e.g
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{
detailViewController *dv = (detailViewController*)segue.destinationViewController;
dv.attribute1 = dataFromTable;
}
例如,如果您在数组中有表数据,那么执行此操作的简单方法是获取行的行号并将其设置为变量
int rowPressed;
- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath{
rowPressed = indexPath.row;
}
然后在PrepareForSegue
中将数据传递给DetailViewController。使用rowPressed从数据模型中获取相关数据。
希望这有帮助!