帮助 .each.replaceWith() jQuery

发布于 2024-10-31 10:48:16 字数 613 浏览 1 评论 0原文

我只是想编写一些代码，对于特定类的每个元素，我需要找到某个属性，然后替换一个元素。我现在的代码看起来像这样：

<div class="a-photo" alt="*/THE IMG SRC*/"></div>
<div class="a-photo" alt="*/THE IMG SRC*/"></div>
<div class="a-photo" alt="*/THE IMG SRC*/"></div>
<div class="a-photo" alt="*/THE IMG SRC*/"></div>

$('.a-photo').each().replaceWith(function(){
var hash = $(this).attr('alt')
"<div class='a-photo' style='background:url("+ hash +") center no-repeat;></div>"
});

我知道这是不对的，而且它不起作用，但我想不出如何写这个，任何帮助将不胜感激！

编辑

我应该提到元素的数量不是预先确定的。

原文

I am just trying to write some code, for each element of a specific class i need to find a certain attribute, then replace an element. the code i have right now looks like this:

<div class="a-photo" alt="*/THE IMG SRC*/"></div>
<div class="a-photo" alt="*/THE IMG SRC*/"></div>
<div class="a-photo" alt="*/THE IMG SRC*/"></div>
<div class="a-photo" alt="*/THE IMG SRC*/"></div>

$('.a-photo').each().replaceWith(function(){
var hash = $(this).attr('alt')
"<div class='a-photo' style='background:url("+ hash +") center no-repeat;></div>"
});

i know this is not right, and its not working, but i can't think of how to write this, any help would be greatly appreciated!

EDIT

i should mention that the amount of elements is not predetermined.

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不喜欢何必死缠烂打 2024-11-07 10:48:16

我想你想要以下内容：

$('.a-photo').each()(function(){
    $(this).css({'background':'url('+$(this).attr('alt') + ') center no-repeat'});
});

I think you want following:

$('.a-photo').each()(function(){
    $(this).css({'background':'url('+$(this).attr('alt') + ') center no-repeat'});
});

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○闲身 2024-11-07 10:48:16

您正在使用 each().replaceWith 这是一个错误，也是多余的，因为 replaceWith 已经做了同样的事情。尝试：

$('.a-photo').replaceWith(function(){
  var alt = $(this).attr('alt');
  return "<div class='a-photo' style='background:url("+ alt +") center no-repeat;></div>";
});

You are using each().replaceWith which is an error and also superfluous becaue replaceWith already does the same. Try:

$('.a-photo').replaceWith(function(){
  var alt = $(this).attr('alt');
  return "<div class='a-photo' style='background:url("+ alt +") center no-repeat;></div>";
});

回复收藏 0 原文

~没有更多了~