Java 为什么我应该使用 url.openStream 而不是 url.getContent？
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Why should i use url.openStream instead of of url.getContent?
提问by juwens
I would like to retrieve the content of a url. Similar to pythons:
我想检索网址的内容。类似于蟒蛇：
html_content = urllib.urlopen("http://www.test.com/test.html").read()
In examples( java2s.com) you see very often the following code:
在示例（java2s.com）中，您经常会看到以下代码：
URL url = new URL("http://www.test.com/test.html");
String foo = (String) url.getContent();
The Description of getContent is the following:
getContent 的描述如下：
Gets the contents of this URL. This method is a shorthand for: openConnection().getContent()
Returns: the contents of this URL.
In my opinion that should work perfectly fine. Buuut obviously this code doesnt work, because it raises an error:
在我看来，这应该工作得很好。但显然这段代码不起作用，因为它引发了一个错误：
Exception in thread "main" java.lang.ClassCastException: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream cannot be cast to java.lang.String
Obviously it returns an inputStream.
显然它返回一个 inputStream。
So i ask myself: what's the purpose of this function which isn't doing what it is seems to do? And why is no hint for quirks it in the documentation? And why did i saw it in several examples?
所以我问自己：这个功能没有做它似乎做的事情的目的是什么？为什么在文档中没有提示它的怪癖？为什么我会在几个例子中看到它？
Or am i getting this wrong?
还是我弄错了？
The suggested solution (stackoverflow) is to use url.openStream() and then read the Stream.
建议的解决方案（stackoverflow）是使用 url.openStream() 然后读取流。
采纳答案by Dave Webb
As you said, documentation says that URL.getContent()is a shortcut for openConnection().getContent()so we need to look at the documentation for URLConnection.getContent().
至于你说的，文件说URL.getContent()是一条捷径openConnection().getContent()，所以我们需要看的文件URLConnection.getContent()。
We can see that this returns an Objectthe type of which is determined by the the content-typeheader field of the response. This type determines the ContentHandlerthat will be used. So a ContentHandlerconverts data based on its MIME type to the appropriate class of Java Object.
我们可以看到，这会返回一个Object类型由content-type响应的头字段确定的类型。此类型确定ContentHandler将使用的。因此 aContentHandler将基于其 MIME 类型的数据转换为相应的 Java 对象类。
In other words the type of Object you get will depend on the content served. For example, it wouldn't make sense to return a Stringif the MIME type was image/png.
换句话说，您获得的 Object 类型将取决于所提供的内容。例如，String如果 MIME 类型是image/png.
This is why in the example code you link to at java2s.com they check the class of the returned Object:
这就是为什么在您链接到 java2s.com 的示例代码中，它们检查返回对象的类的原因：
try {
URL u = new URL("http://www.java2s.com");
Object o = u.getContent();
System.out.println("I got a " + o.getClass().getName());
} catch (Exception ex) {
System.err.println(ex);
}
So you can say String foo = (String) url.getContent();if you know your ContentHandlerwill return a String.
所以你可以说String foo = (String) url.getContent();如果你知道你ContentHandler会返回一个String.
There are default content handlers defined in the sun.net.www.contentpackage but as you can see they are returning streams for you.
sun.net.www.content包中定义了默认的内容处理程序，但正如您所见，它们正在为您返回流。
You could create your own ContentHandlerthat does return a Stringbut it will probably be easier just to read the Stream as you suggest.
您可以创建自己的ContentHandler返回 a 的内容，String但按照您的建议阅读 Stream 可能会更容易。
回答by Martijn Courteaux
You misunderstand what "Content" means. You expected it to return a String containing the HTML, but it returns a HttpInputStream. Why? Because the requested URL is a html webpage. Another valid URL might be http://www.google.com/logo.png. This URL doesn't contain String content. It is an image.
您误解了“内容”的含义。您希望它返回一个包含 HTML 的字符串，但它返回一个 HttpInputStream。为什么？因为请求的 URL 是一个 html 网页。另一个有效的 URL 可能是http://www.google.com/logo.png. 此 URL 不包含字符串内容。它是一个图像。
回答by prunge
You can use Guava's Resources.toString(URL, Charset)method to more easily read a URL to a string.
您可以使用Guava的Resources.toString(URL, Charset)方法更轻松地读取 URL 到字符串。
