Question

我正在使用SwiftyJSON从外部网站下载图像并在表格视图中显示它们。我想点击一个特定的图像，让它打开一个外部URL。怎么做到这一点？

基本上我通过按下按钮（如下所示），我想通过点击图像来做。

    override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
    let cell = tableView.dequeueReusableCellWithIdentifier("cell", forIndexPath: indexPath) as! myCell

    let str = TableData[indexPath.row].thumbnail
    let imagePath = (str.substringToIndex(str.startIndex.advancedBy((str.characters.count)-11)))+".png" //Removing "-100x90.png" from each thumbnail path and appending extension .png later on to fetch original sized image

    cell.myImageView.kf_setImageWithURL(NSURL(string: imagePath)!)
    cell.myLabel.text = String(TableData[indexPath.row].date) + ":" + String(TableData[indexPath.row].title)
    print(TableData[indexPath.row].permalink)
    cell.myButton.tag = indexPath.row
    cell.myButton.addTarget(self, action: "openURL:", forControlEvents: UIControlEvents.TouchUpInside)
    print(cell.myButton.tag) 
    return cell
}


func openURL(sender:UIButton){
print("Button Pressed")
UIApplication.sharedApplication().openURL(NSURL(string: TableData[sender.tag].permalink)!)
}

Answer 1

将UITapGestureRecognizer添加到单元格中的图像视图中：

let tapRecognizer = UITapGestureRecognizer(target: self, action: "openURL:")
cell.myImageView.addGestureRecognizer(tapRecognizer)

Answer 2

它适用于以下代码：

    let singleTap = UITapGestureRecognizer(target: self, action:"tapDetected:")
    singleTap.numberOfTapsRequired = 1
    cell.myImageView.userInteractionEnabled = true
    cell.myImageView.tag = indexPath.row
    cell.myImageView.addGestureRecognizer(singleTap)

func tapDetected(sender: UITapGestureRecognizer) {
    UIApplication.sharedApplication().openURL(NSURL(string: TableData[(sender.view!.tag)].link)!)
}

如何使用SWIFT2.0单击图像打开外部链接？

2 个答案: