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我有以下 XML：

<item p4:stage="Ways and Means resolution" xmlns:p4="urn:services.parliament.uk-bills.ext"><guid isPermaLink="true">http://services.parliament.uk/bills/2013-14/antisocialbehaviourcrimeandpolicingbill.html</guid><link>http://services.parliament.uk/bills/2013-14/antisocialbehaviourcrimeandpolicingbill.html</link><category>Commons</category><category>Government Bill</category><title>Anti-social Behaviour, Crime and Policing</title><description>To make provision about anti-social behaviour, crime and disorder, including provision about recovery of possession of dwelling houses; to make provision amending the Dangerous Dogs Act 1991, Schedules 7 and 8 to the Terrorism Act 2000 and the Extradition Act 2003; to make provision about firearms and about forced marriage; to make provision about the police, the IndependentPolice Complaints Commission and the Serious Fraud Office; to make provision about criminal justice and court fees; and for connected purposes.</description><a10:updated>2013-06-05T17:42:59+01:00</a10:updated></item>

我可以<item></item>使用它解析所有内容：

$(xml).find('item').each(function() {
    var stage = $(this).find('title').text();
    console.log(stage);
});

但我正在查看里面的内容：

<item....>  for example,

我需要把里面的东西拉出来：

 <p4:stage></p4:stage>

我怎么把这个拉出来？

先感谢您

4

1 回答 1

3

无效的 XML中没有<p4:stage></p4:stage>，p4:stage是 item 元素上的属性

$(xml).find('item').each(function() {
    var p4 = $(this).attr('p4:stage');
    console.log(p4);
});

小提琴

于 2013-06-05T21:05:22.070 回答