SAX error: incompatible types: String cannot be converted to InputSource

Question

相关代码； barfs 实例化 SAXSource:

TransformerFactory factory = TransformerFactory.newInstance();
XMLReader xmlReader = XMLReaderFactory.createXMLReader("org.ccil.cowan.tagsoup.Parser");
Source input = new SAXSource(xmlReader, "http://books.toscrape.com/");
Result output = new StreamResult(System.out);
factory.newTransformer().transform(input, output);

JavaDoc 的 say:

public SAXSource(XMLReader reader,
         InputSource inputSource)

Create a SAXSource, using an XMLReader and a SAX InputSource. The Transformer or SAXTransformerFactory will set itself to be the reader's ContentHandler, and then will call reader.parse(inputSource).

查看InputSource显示：

InputSource(InputStream byteStream)
Create a new input source with a byte stream.
InputSource(Reader characterStream)
Create a new input source with a character stream.

因此，对于 example，这需要在 html 中读取 InputStream??

的字符流

tagsoup 是否更好地用于身份转换？但是，怎么办？

Answer 1

您可以像这样访问从 URL 后面的资源中读取的 InputStream：

InputStream i = new URL("http://...").openConnection().getInputStream();

然后您可以使用 i 作为您的 SAXSource。

Answer 2

有一个构造函数 https://docs.oracle.com/javase/8/docs/api/org/xml/sax/InputSource.html#InputSource-java.lang.String- 接受系统 ID，例如一个 URL 所以你可以使用 Source input = new SAXSource(xmlReader, new InputSource("http://books.toscrape.com/"));.

SAX error: incompatible types: String cannot be converted to InputSource

SAX error: incompatible types: String cannot be converted to InputSource

java

xml

xslt

tag-soup

saxon