10

我有以下html：

<div class="threeimages" id="txtCss">  
<a>   
       <img alt="Australia" src="/Images/Services%20button_tcm7-9688.gif"/>
</a>        
    <div class="text" id="txtLink">            
        <h2>
            <a href="/partnerzone/downloadarea/school-information/australia/index.aspx">Australia</a>
        </h2>            
        <p>Land of the sunshine!</p>        
    </div>
</div>

现在，如果您看到 div ID“txtLink”中有 href，即澳大利亚

我希望在页面运行时将相同的href值复制到div ID“txtCss”的上面标签中，我的意思是当我的页面显示时，我的html将如下所示：

<div class="threeimages" id="txtCss">  
<a href="/partnerzone/downloadarea/school-information/australia/index.aspx">   
       <img alt="Australia" src="/Images/Services%20button_tcm7-9688.gif"/>
</a>        
    <div class="text" id="txtLink">            
        <h2>
            <a href="/partnerzone/downloadarea/school-information/australia/index.aspx">Australia</a>
        </h2>            
        <p>Land of the sunshine!</p>        
    </div>
</div>

请为上述问题提出一些代码

4

4 回答 4

38

这是不使用任何库的最短答案，并且只适用于您想要的东西

var tc = document.getElementById("txtCss");
var ary = tc ? tc.getElementsByTagName("a") : [];
if(ary.length >= 2)
    ary[0].href = ary[1].href;
于 2009-05-20T11:12:41.753 回答
11

更新
这里没有 jquery 的答案：https
://stackoverflow.com/a/887348/11333 早在 2009 年，使用 jquery 是完全可以接受的:)

使用以下内容创建一个 js 文件：

$(document).ready(function() {
    $('.threeimages').each(function(){
    $(this).DupeHref();
    });
});

jQuery.fn.DupeHref =  function(){       
    var target = $(this).find(".text h2 a").attr("href");
    $(this).find("a").attr("href", target);
}

在你的 html 中引用 jquery 和这个 javascript 文件。像这样的东西：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
        <title>Test</title>         
    </head>
    <body>
        <div class="threeimages">  
            <a>   
                <img alt="Australia" src="/Images/Services%20button_tcm7-9688.gif"/>
            </a>        
            <div class="text">            
                <h2>
                    <a href="/partnerzone/downloadarea/school-information/australia/index.aspx">Australia</a>
                </h2>            
                <p>Land of the sunshine!</p>        
            </div>
        </div>
        <div class="threeimages">  
                     <a>   
                <img alt="Belgium" src="/Images/Services%20button_tcm7-9689.gif"/>
            </a>        
            <div class="text">            
                <h2>
                        <a href="/partnerzone/downloadarea/school-information/belgium/index.aspx">Belgium</a>
                </h2>            
                <p>Land of beer and food!</p>        
            </div>
        </div>
        <script src="./content/js/jquery-1.2.6.min.js" type="text/javascript"></script>
        <script src="./content/js/dupetargets.js" type="text/javascript"></script>
    </body>
</html>
于 2009-05-20T11:06:37.103 回答
2

这是一个简单易懂的代码：

 <html>
  <head>
   <script language="javascript">  
  function simpleChangeURL(){
        var anchors = document.getElementsByTagName('a');
        if(anchors != null & anchors.length > 0){
            anchors[0].href = anchors[1].href;  
        }   
    }

    function simpleChangeByAustraliaURL()
    {
        var anchors = document.getElementsByTagName('a');
        var images = document.getElementsByTagName('img');
        var imageNeeded;
        var anchorNeeded;
        if(images != null){
            for( var i = 0; i < images.length ; i++){
                if (images[i].alt == 'Australia') imageNeeded = images[i];
            }
        }
        if(anchors != null){
            for( var j = 0; j < anchors.length ; j++){
                if (anchors[j].firstChild.data == 'Australia') anchorNeeded = anchors[j];
            }
        }
        if(imageNeeded != null && anchorNeeded!= null){
    var imageAnchor = imageNeeded.parentNode;
    imageAnchor.href = anchorNeeded;

}

    }


    </script>
    </head>

    <body>
    <div class="threeimages" id="txtCss">  
    <a>   
           <img alt="Australia" src="/Images/Services%20button_tcm7-9688.gif"/>
    </a>        
        <div class="text" id="txtLink" >            
            <h2>
                    <a href="/partnerzone/downloadarea/school-information/australia/index.aspx">Australia</a>
            </h2>            
            <p>Land of the sunshine!</p>        
        </div>
    </div>
    <script> simpleChangeByAustraliaURL();</script>
    </body>
    </html>
于 2009-05-20T12:38:52.470 回答
1

您可以使用：

var txtLink = document.getElementById('txtLink');
var a = txtLink.getElementsByTagName('a');
if (a != null && a.length > 0) {
    var setLink = txtLink.parentNode.getElementsByTagName('a');
    if (setLink != null && setLink.length > 0) {
        setLink[0].href = a[0].href;
    }
}

我认为这应该工作..

于 2009-05-20T11:10:38.653 回答