向jetty添加多个资源目录 [英] Add more than one resource directory to jetty

问题描述

希望与Jetty使用多个静态目录。当服务器运行时：

  http：// localhost：8282 / A 
 http：// localhost：8282 / B 
 http：// localhost：8282 / C 
  

• A被放置在X / V / A中


• B放在Q / Z / B中


• C放在P / T / C

以下失败：

  ResourceHandler resource_handler = new ResourceHandler（）; 
 resource_handler.setWelcomeFiles（new String [] {index.html}）; 
 resource_handler.setResourceBase（HTML_SITE）; 
 
 ResourceHandler resource_handler1 = new ResourceHandler（）; 
 resource_handler1.setWelcomeFiles（new String [] {index.html}）; 
 resource_handler1.setResourceBase（HTML_CLIENTZONE_SITE）; 
 
 //部署引擎
 WebAppContext webapp = new WebAppContext（）; 
 
 String dir = System.getProperty（user.dir）; 
 webapp.setResourceBase（getWebAppPath（））; 
 webapp.setContextPath（/）; 
 
 
 HandlerList handlers = new HandlerList（）; 
 handlers.setHandlers（new Handler [] {resource_handler，resource_handler1，webapp，new DefaultHandler（）}）; 
 server.setHandler（处理程序）; 
  

如何添加多个静态资源目录？

解决方案

自6.1.12以来，通过使用ResourceCollection到WebAppContext的基础资源来支持这一点：

 服务器服务器=新服务器（8282）; 
 WebAppContext context = new WebAppContext（）; 
 context.setContextPath（/）; 
 ResourceCollection resources = new ResourceCollection（new String [] {
project / webapp / folder，
/ root / static / folder / A，
/ root /静态/文件夹/ B，
}）; 
 context.setBaseResource（resources）; 
 server.setHandler（context）; 
 server.start（）; 
  

要随后打开一个文件，请使用ServletContext（例如WebAppContext），它可能是界面定义，如：

  / ** 
 * servlet上下文。 
 * / 
 public default InputStream open（ServletContext context，String filename）{
 String f = System.getProperty（file.separator）+ filename; 
 return context.getResourceAsStream（f）; 
} 
  

如：

< pre class =lang-java prettyprint-override> InputStream in = open（context，filename.txt）;


这将打开 filename.txt 存在于一个给定的目录中。请注意，getResourceAsStream将返回 null ，而不是抛出异常，所以最好检查它：

  public default InputStream validate（InputStream in，String filename）
 throws FileNotFoundException {
 if（in == null）{
抛出新的FileNotFoundException（filename）; 
} 
 
返回; 
} 
  

然后你可以更新打开方法如下：

  return validate（context.getResourceAsStream（filename），filename） ; 
  

Looking to use multiple static directories with Jetty. When the server runs:

  http://localhost:8282/A
  http://localhost:8282/B 
  http://localhost:8282/C

• A is placed in X/V/A
• B is placed in Q/Z/B
• C is placed in P/T/C

The following failed:

    ResourceHandler resource_handler = new ResourceHandler();
    resource_handler.setWelcomeFiles(new String[]{"index.html"});
    resource_handler.setResourceBase(HTML_SITE);

    ResourceHandler resource_handler1 = new ResourceHandler();
    resource_handler1.setWelcomeFiles(new String[]{"index.html"});
    resource_handler1.setResourceBase(HTML_CLIENTZONE_SITE);

    // deploy engine
    WebAppContext webapp = new WebAppContext();

    String dir = System.getProperty("user.dir");
    webapp.setResourceBase(getWebAppPath());
    webapp.setContextPath("/");


     HandlerList handlers = new HandlerList();
    handlers.setHandlers(new Handler[]{resource_handler,resource_handler1 ,webapp,  new DefaultHandler()});
    server.setHandler(handlers);

How can I add more than one static resource directory?

解决方案

Since 6.1.12, this is supported by using a ResourceCollection to the WebAppContext's base resource:

Server server = new Server(8282);
WebAppContext context = new WebAppContext();
context.setContextPath("/");
ResourceCollection resources = new ResourceCollection(new String[] {
    "project/webapp/folder",
    "/root/static/folder/A",    
    "/root/static/folder/B",    
});
context.setBaseResource(resources);
server.setHandler(context);
server.start();

To subsequently open a file, use the ServletContext (e.g., WebAppContext), which could be part of an interface definition, such as:

  /**
   * Opens a file using the servlet context.
   */
  public default InputStream open( ServletContext context, String filename ) {
    String f = System.getProperty( "file.separator" ) + filename;
    return context.getResourceAsStream( f );
  }

Such as:

  InputStream in = open( context, "filename.txt" );

This will open filename.txt if it exists in one of the given directories. Note that getResourceAsStream will return null, rather than throw an exception, so it's a good idea to check for it:

  public default InputStream validate( InputStream in, String filename )
    throws FileNotFoundException {
    if( in == null ) {
      throw new FileNotFoundException( filename );
    }

    return in;
  }

Then you can update the open method as follows:

return validate( context.getResourceAsStream( filename ), filename );

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