java Programming Glossary: inputsource

http://stackoverflow.com/questions/139076/how-to-pretty-print-xml-from-java

import org.w3c.dom.Document import org.xml.sax.InputSource import org.xml.sax.SAXException import javax.xml.parsers.DocumentBuilder.. DocumentBuilder db dbf.newDocumentBuilder InputSource is new InputSource new StringReader in return db.parse is catch.. db dbf.newDocumentBuilder InputSource is new InputSource new StringReader in return db.parse is catch ParserConfigurationException..

Looking for a CSS Parser in java

http://stackoverflow.com/questions/1513587/looking-for-a-css-parser-in-java

import org.w3c.css.sac.InputSource import org.w3c.dom.css.CSSStyleSheet import org.w3c.dom.css.CSSRuleList.. stderr to the log file as well else return rtn InputSource source new InputSource new InputStreamReader stream CSSOMParser.. file as well else return rtn InputSource source new InputSource new InputStreamReader stream CSSOMParser parser new CSSOMParser..

Make DocumentBuilder.parse ignore DTD references

http://stackoverflow.com/questions/155101/make-documentbuilder-parse-ignore-dtd-references

new EntityResolver @Override public InputSource resolveEntity String publicId String systemId throws SAXException.. IOException if systemId.contains foo.dtd return new InputSource new StringReader else return null I found that simply returning..

How to read XML response from a URL in java?

http://stackoverflow.com/questions/2310139/how-to-read-xml-response-from-a-url-in-java

myReader.setContentHandler handler myReader.parse new InputSource new URL url .openStream or if you prefer DOM DocumentBuilderFactory..

JAXB: How to ignore namespace during unmarshalling XML document?

http://stackoverflow.com/questions/277502/jaxb-how-to-ignore-namespace-during-unmarshalling-xml-document

source XMLReader reader new NamespaceFilterXMLReader InputSource is new InputSource fr SAXSource ss new SAXSource reader is return.. reader new NamespaceFilterXMLReader InputSource is new InputSource fr SAXSource ss new SAXSource reader is return unmarshaller.unmarshal..

Java URLConnection Timeout

http://stackoverflow.com/questions/3163693/java-urlconnection-timeout

true InputStream inStream urlConn.getInputStream InputSource input new InputSource inStream And I am catching the SocketTimeoutException... inStream urlConn.getInputStream InputSource input new InputSource inStream And I am catching the SocketTimeoutException. Thanks..

How do I read a resource file from a Java jar file?

http://stackoverflow.com/questions/403256/how-do-i-read-a-resource-file-from-a-java-jar-file

this xr.setErrorHandler this xr.parse new InputSource new FileReader filename What's wrong with using this technique..

how to retrieve element value of XML using Java?

http://stackoverflow.com/questions/4076910/how-to-retrieve-element-value-of-xml-using-java

Document document builder.parse new InputSource new StringReader xml Element rootElement document.getDocumentElement..

In Java, how do I parse XML as a String instead of a file?

http://stackoverflow.com/questions/562160/in-java-how-do-i-parse-xml-as-a-string-instead-of-a-file

DocumentBuilder builder factory.newDocumentBuilder InputSource is new InputSource new StringReader xml return builder.parse.. builder factory.newDocumentBuilder InputSource is new InputSource new StringReader xml return builder.parse is also see this similar..

Order of XML attributes after DOM processing

http://stackoverflow.com/questions/726395/order-of-xml-attributes-after-dom-processing

Source src new SAXSource sp.getXMLReader new InputSource input.getAbsolutePath String resultFileName input.getAbsolutePath..

Simplest way to query XML in Java

http://stackoverflow.com/questions/807418/simplest-way-to-query-xml-in-java

XPathFactory.newInstance XPath xpath xpathFactory.newXPath InputSource source new InputSource new StringReader xml String status xpath.evaluate.. XPath xpath xpathFactory.newXPath InputSource source new InputSource new StringReader xml String status xpath.evaluate resp status..