javascript 如何从 JQuery 的 AJAX 回调函数显示整个 XML？
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How to display entire XML from JQuery's AJAX callback function?
提问by limc
I have this simple code to query against the web service:-
我有这个简单的代码来查询网络服务：-
$.get( url ,
function(xml) {
var hello = $(xml).find("hello").text();
...
alert(xml); // displays [object XMLDocument]
alert($(xml)); // displays [object Object]
}
);
This works fine, but I'm interested to see entire XML structure from the callback function for debugging purpose. I tried a few things, but I couldn't get it to display the XML. What I want to see is something like this:-
这工作正常，但我有兴趣从回调函数中查看整个 XML 结构以进行调试。我尝试了一些东西，但我无法让它显示 XML。我想看到的是这样的：-
<stuff>
<hello>bear</hello>
</stuff>
Any clue? Thanks.
有什么线索吗？谢谢。
回答by Scoobler
If you are using firebug in firefox, but may also work in IE8 or chrome, you can try:
如果您在 firefox 中使用 firebug，但也可能在 IE8 或 chrome 中工作，您可以尝试：
console.dirxml(xml)OR
console.dir(xml)OR
console.log(xml)
console.dirxml(xml)或
console.dir(xml)或
console.log(xml)
In IE8 press F12 to open up the developers console or you may get the error saying the normal browser doesn't know what console is (after F12 it should).
在 IE8 中按 F12 打开开发者控制台，否则你可能会收到错误消息，说普通浏览器不知道控制台是什么（在 F12 之后应该）。
Alternatively you could use prettyPrintto display the object. Or take a look at the answer to Is there an equivalent for var_dump (PHP) in Javascript?
或者，您可以使用PrettyPrint来显示对象。或者看看在 JavaScript 中是否有 var_dump (PHP) 的等价物的答案？
回答by jAndy
You would need a little trickery. Introduce a new element, which wraps your XML and output the .html()from that. Like
你需要一点技巧。引入一个新元素，它包装您的 XML 并.html()从中输出。喜欢
var fakexml = "<stuff><hello>bear</hello></stuff>";
alert($('<debug>').append(fakexml).html());
or use .wrapAll()
或使用 .wrapAll()
alert($(fakexml).wrapAll('<debug>').parent().html());
