问题描述
我在IOS5应用中为我的表格单元设置了一个手势识别器:
I have set up a gesture recognizer for my table cell in my IOS5 app:
UITapGestureRecognizer* oneFingerDoubleTap = [[UITapGestureRecognizer alloc]initWithTarget:self action:@selector(cellOneFingerDoubleTap:)];
oneFingerDoubleTap.numberOfTapsRequired = 2;
[cell addGestureRecognizer:oneFingerDoubleTap];
并实现了处理程序方法:
And implemented handler method:
- (void)cellOneFingerDoubleTap:(id) sender
{
NSLog(@"taptap");
}
一切正常。我的问题是我无法通过窃听细胞传递细胞或其他一些数据。正如我所见(id)发送者是UITapGestureRecognizer本身。
It works fine. My problem is that I can not pass the cell was tapped or some other data with the tapped cell. As I see (id)sender is the UITapGestureRecognizer itself.
我的问题是:如何在处理程序方法(cellOneFingerDoubleTap)中获取tapped单元格?如何在处理程序方法中获取tapped单元格的索引?
My question is: how can I get the tapped cell in the handler method (cellOneFingerDoubleTap)? How can I get in the handler method the index of the tapped cell?
谢谢!
推荐答案
如果你从传递的手势识别器中获取视图
在您的 cellOneFingerDoubleTap:
方法中,然后您将获得已被点击的单元格。类似于:
If you grab the view
from the gesture recogniser that is passed in to you on your cellOneFingerDoubleTap:
method, then you'll get the cell that's been tapped. Something like:
- (void)cellOneFingerDoubleTap:(UIGestureRecognizer*)recognizer {
UITableViewCell *cell = (UITableViewCell*)recognizer.view;
}
[我只假设一个单元格你的意思是 UITableViewCell
顺便说一下]
[I'm just assuming by a "cell" you mean a UITableViewCell
by the way]
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