Question

代码：

var links = document.getElementsByTagName('link');
var links_length = links.length;

for (var i = 0; i < links_length; i++)
{
    if (links[i].getAttribute('rel').indexOf('test') !== -1)
    {
        //blabla
        // but now I want to target for example... HREF attribute of link that got test in it's rel attribute... so

        if (links[i].getAttribute('href').indexOf('test2') !== -1)
        {
            //blabla2

            // but it doesn't work
            // it won't give me href with test2 of the link with rel attribute with test. The code give my last link with href of test2... dunno why. Any ideas?
        }
    }
}

为什么我无法与rel='test'和href='test2'建立正确的链接？ href属性迭代应仅针对rel='test'的链接，对吗？

HTML：

<a href="test2" rel="test">Proper Test</a> // I want to grab this one
<a href="testing" rel="testing">Test</a>
<a href="testing" rel="testing">Test</a>
<a href="test2" rel="besting">Test</a>
<a href="testing" rel="test">Test</a>

Answer 1

修正了一些js错误并对html做了假设。只需使用console.log，alert或控制台调试器逐步进行故障排除，您就可以捕获这些内容。

＆＃13;

var links = document.getElementsByTagName('a');

for (var i = 0; i < links.length; i++) {
  if (links[i].getAttribute('rel') == 'test') {

    if (links[i].getAttribute('href') == 'test2') {

      console.log("link index:" + i );

    }
  }
}

＆＃13;

<a href="test2" rel="test">Test</a>
<a href="best" rel="testing">Test</a>
<a href="thetest2" rel="testing">Test</a>
<a href="test2" rel="besting">Test</a>
<a href="thetest2" rel="testing">Test</a>

＆＃13;

Answer 2

通过查看您的评论，我认为您已经陷入了javascript关闭。通过这个循环很简单，你只需要在满足条件的地方打破。

var links = document.getElementsByTagName('link');
var mylink = '';

for (var i = 0; i < links.length; i++)
{
    if (links[i].getAttribute('rel').indexOf('test') !== -1)
    {
       if (links[i].getAttribute('href').indexOf('test2') !== -1)
        {
            mylink = links[i];
            break; //If you need the first one. else create an array and keep pushing the ietms.
        }
    }
}

console.log(mylink);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="test" href="www.test2.com">
<link rel="test1"  href="www.test2.com">
<link rel="test2"  href="www.test2.com">
<link rel="test3"  href="www.test2.com">

迭代数组不能给出正确的结果

2 个答案: