使用jquery,如何选择div内的所有图像来更改其来源

时间:2017-08-18 07:28:41

标签: jquery image

我有以下div,我知道DIV的选择器ID。

<div class="event">
  <img src="/Content/Images/Icons/calendar1.png">
  <img src="/Content/Images/Icons/calendar2.png">
  <img src="/Content/Images/Icons/calendar3.png">
  <img src="/Content/Images/Icons/calendar4.png">
  <img src="/Content/Images/Icons/calendar5.png">
  <img src="/Content/Images/Icons/calendar6.png">
</div>

我需要一些东西来找到我所拥有的div中的所有图像选择器,所以我可以将每个图像的源更改为新图像。

6 个答案:

答案 0 :(得分:4)

使用此代码检索图片网址:

$('.event img').each(function(){
  alert($(this).attr('src'));
});

如果您想更改图片网址,请使用以下网址:

var inc = 1;
$('.event img').each(function(){
  $(this).attr('src','path of image/imagename'+inc+'.imageextension');
  inc++;
});

答案 1 :(得分:3)

您可以一起尝试子选择器和attr回调。

&#13;
&#13;
  $(".event img").attr("src", function() { // or id
      return "Something/" + $(this).attr("src");
 });
   console.log($(".event").html());
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="event">
  <img src="/Content/Images/Icons/calendar1.png">
  <img src="/Content/Images/Icons/calendar2.png">
  <img src="/Content/Images/Icons/calendar3.png">
  <img src="/Content/Images/Icons/calendar4.png">
  <img src="/Content/Images/Icons/calendar5.png">
  <img src="/Content/Images/Icons/calendar6.png">
</div>
&#13;
&#13;
&#13;

答案 2 :(得分:3)

你可以循环遍历每个元素:

i代表每个元素的索引,我已经添加了这个你想要的东西

imgurl1 imgulr2 imgurl3

等等

&#13;
&#13;
$(".event img").each(function(i, x) {
  $(this).attr("src", "/Content/Images/Icons/NewURL" + (i + 1) + ".png")
  console.log($(this).attr("src"))
})
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="event">
  <img src="/Content/Images/Icons/calendar1.png">
  <img src="/Content/Images/Icons/calendar2.png">
  <img src="/Content/Images/Icons/calendar3.png">
  <img src="/Content/Images/Icons/calendar4.png">
  <img src="/Content/Images/Icons/calendar5.png">
  <img src="/Content/Images/Icons/calendar6.png">
</div>
&#13;
&#13;
&#13;

答案 3 :(得分:1)

您可以按照代码段返回img元素列表,

$('.event').find('img');

答案 4 :(得分:1)

将所有内容更改为相同:

$(".event>img").attr("src","/New/Image.jpg");

将所有内容更改为:

$(".event>img").each(function(index){
    $(this).attr("src",variable);
});

答案 5 :(得分:1)

使用数组选择器为此:img[0] , img[1].Attr方法("src","url"); 或者你可以给他们的课程来选择每个img !!