从流输入中解析没有根元素的XML片段列表 [英] Parse a list of XML fragments with no root element from a stream input

问题描述

在Java中使用SAX api来解析流输入中没有根元素的XML片段列表是否可行？

Is it feasible in Java using the SAX api to parse a list of XML fragments with no root element from a stream input?

我尝试解析这样的XML但是得到了

I tried parsing such an XML but got a

org.xml.sax.SAXParseException: The markup in the document following the root element must be well-formed.

甚至在触发endDocument事件之前。

before even the endDocument event was fired.

我不想解决明显但笨拙的解决方案，如预先添加自定义根元素或使用缓冲片段解析。

I would like not to settle with obvious but clumsy solutions as "Pre-append a custom root element or Use buffered fragment parsing".

我使用的是Java 1.6的标准SAX API。如果有人想知道的话，SAX工厂已经设置了Validating（假）。

I am using the standard SAX API of Java 1.6. The SAX factory had setValidating(false) in case anyone wondered.

推荐答案

首先，最重要的是，你的内容是解析不是XML文档。
来自 XML规范：

First, and most important of all, the content you are parsing is not an XML document. From the XML Specification:

[定义：只有一个元素，称为根或文档元素，其中没有一部分出现在内容中任何其他元素。]

[Definition: There is exactly one element, called the root, or document element, no part of which appears in the content of any other element.]

现在，关于用SAX解析这个问题 - 尽管你说的是笨拙 - 我建议以下方法：

Now, as to parsing this with SAX - in spite of what you said about clumsiness - I'd suggest the following approach:

Enumeration<InputStream> streams = Collections.enumeration(
    Arrays.asList(new InputStream[] {
        new ByteArrayInputStream("<root>".getBytes()),
        yourXmlLikeStream,
        new ByteArrayInputStream("</root>".getBytes()),
    }));

SequenceInputStream seqStream = new SequenceInputStream(streams);

// Now pass the `seqStream` into the SAX parser.

使用 SequenceInputStream 是将多个输入流连接成单个流的便捷方式。它们将按照传递给构造函数的顺序读取（或者在这种情况下 - 由 Enumeration 返回）。

将它传递给你的SAX解析器，你就完成了。

Pass it to your SAX parser, and you are done.

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