回调到.delay() [英] Callback to .delay()

问题描述

我有一个$image，我是.fadeIn和.fadeOut，然后是.remove.这是我的代码:

I have an $image that I .fadeIn and .fadeOut, and then .remove after .fadeOut completes. This is my code:

$image
   .fadeIn()
   .fadeOut(function() {
      $(this).remove();
   });

我想在.fadeOut之后添加.delay，并且仅在.delay完成后添加.remove $image.我已经尝试过:

I want to add a .delay after .fadeOut, and .remove the $image only once .delay has completed. I have tried:

$image
   .fadeIn()
   .fadeOut()
   .delay(1000, function() {
      $(this).remove();
   });

问题是.delay不不接受回调函数.我如何.remove图片作为.delay的回调?

The problem is that .delay doest not accept a callback function. How can I .remove the picture as a callback to .delay?

推荐答案

您可以使用 queue()安排您自己的函数在delay()完成后运行的方法:

You can use the queue() method to schedule your own function to run after delay() completes:

$image.fadeIn()
      .fadeOut()
      .delay(1000)
      .queue(function(next) {
          $(this).remove();
          next();
      });

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