Code for validating xml against xsd in java
Here is the sample XML file: In the jar file, the Xsd Gen constructors are default package access specifies, so you can not create the instance of this class outside of the package, you need to modify the source code of Xsd Gen constructors to public access specifies to create the instance of this class. You will find the source code of the Xsd Gen class in the following location: " Ok, no big deal I figure I can declare the xsd file in my xml, but it won't let me because when it starts parsing it says: Error: URI=file: C:/Documents and Settings/.../Line=4: Element type "input" is not declar ed.
programming forums Java Mobile Certification Databases Caching Books Engineering Micro Controllers OS Languages Paradigms IDEs Build Tools Frameworks Application Servers Open Source This Site Careers Other all forums Hi , I am using XML SPY version5 Professional Edition , evaluation copy in my project development. For this type(any Type) , I can add either test between the elements or child elements itself.
This article shows five ways of how to configure different Java APIs (including DOM, SAX, dom4j and XOM) using JAXP 1.3 for checking and validating XML with DTD and Schema(s).
To report errors, it is necessary to provide an Error Handler to the underlying implementation.
Hello, I am trying to validate an xml file against an xsd in java 1.4.
The code below works flawlessly in java 1.5 but in 1.4 I can't set Attributes.