在UITapGestureRecognizer处理程序中获取tapped单元格 [英] Get tapped cell in UITapGestureRecognizer handler

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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