Standalone validation for xml using xsd system

When we put xsd in xsi: schemaLocation, download xsd from the Internet and check xml?

I want to do a standalone check by downloading xsd, if so, how can I do this?

I know how to do a standalone DTD check, but not sure about xsd.

For DTD, I am currently doing the following:

<!DOCTYPE struts-config SYSTEM "struts-config_1_1.dtd">

Can I do something like this?

WITH

<!-- beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="http://www.springframework.org/schema/beans 
           http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
           http://www.springframework.org/schema/context
           http://www.springframework.org/schema/context/spring-context-3.0.xsd"-->

before

<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="http://www.springframework.org/schema/beans 
                           file://spring-beans-3.0.xsd
                           http://www.springframework.org/schema/context
                           file://spring-context-3.0.xsd">

EDIT:

Looks like I'm wrong. spring is only sent to the Internet when it does not find dtd / xsd in the jar file. I pulled out the internet plug and saw that the application was working fine.

In addition, the source: https://forum.hibernate.org/viewtopic.php?f=1&t=949031

Sorry for the problem and the wrong question.