How do I create an org.xml.sax.InputSource from a String?

JavaXml

Java Problem Overview

I am following a guide, and it gives me the following code:

InputSource inputSource = new InputSource(new FileInputStream(new File("/path/to/xml/file.xml"))));

What I would like to know, is how I can still create an org.xml.sax.InputSource, but instead of reading the content of a file, use a String variable that I already have.

Java Solutions

Solution 1 - Java

Use a StringReader instead of a FileInputStream.

See the documentation for StringReader

example:

InputSource inputSource = new InputSource( new StringReader( myString ) );

Solution 2 - Java

InputSource inputSource = new InputSource(new java.io.StringReader(string)); if it is org.xml.sax.InputSource.

Categories
Recommended Java Solutions
Recommended Xml Solutions

Previous Solution:

F#: String.Join and |> operator

F#F# Interactive

Next Solution:

PHP header() redirect with POST variables

PhpHtmlPostHeader

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionChigginsView Question on Stackoverflow
Solution 1 - JavaGavin HView Answer on Stackoverflow
Solution 2 - JavakhachikView Answer on Stackoverflow