java svg image xlink,获取SVG< image>的“xlink：href”属性元素在HTML DOM中动态地使用JS...

I have a construction:

I want to get pic.jpg url and I need to begin from the most outer div, not exactly from the source element:

var div = document.getElementById("div");

var svg = div.getElementsByTagNameNS('http://www.w3.org/2000/svg', 'svg')[0];

var img = svg.getElementsByTagNameNS('http://www.w3.org/2000/svg', 'image')[0];

var url = img.getAttribute('xlink:href'); // Please pay attention I do not use getAttributeNS(), just usual getAttribute()

alert(url); // pic.jpg, works fine

My question is what is the right way to get such kind of attributes from element like SVG and its children?

Because before I tried to do this way and it also worked fine in Chrome (I didn't try other browsers):

var svg = div.getElementsByTagName('svg')[0]; // I do not use NS

var img = svg.getElementsByTagName('image')[0];

var url = img.getAttribute('xlink:href'); // and do not use getAttributeNS() here too

alert(url); // pic.jpg, works fine

But when I tried to use getAttributeNS() I got blank result:

var svg = div.getElementsByTagNameNS('http://www.w3.org/2000/svg', 'svg')[0];

var img = svg.getElementsByTagNameNS('http://www.w3.org/2000/svg', 'image')[0];

// Please pay attention I do use getAttributeNS()

var url = img.getAttributeNS('http://www.w3.org/1999/xlink', 'xlink:href');

alert(url); // but I got black result, empty alert window

解决方案

The correct usage is getAttributeNS('http://www.w3.org/1999/xlink', 'href');