In web.xml:
<servlet>
<servlet-name>imageReceiver</servlet-name>
<servlet-class>xxx.ImageReceiver</servlet-class>
<init-param>
<param-name>imagePath</param-name>
<param-value>/the/place</param-value><!-- or C:\the\place -->
</init-param>
</servlet>
And in the servlet:
public void init(ServletConfig servletConfig) throws ServletException{
imagePath = servletConfig.getInitParameter("imagePath");
if (!imagePath.endsWith(File.separator)) imagePath += File.separator
try {
File imageFolder = new File(imagePath);
if (!File.exists()) Util.mkdirRecursive(imagePath);
} catch ... whatever you want to do if something fails
}
El martes 3 de abril de 2012 03:21:06 UTC+2, Sam W escribió:
>
> Hello,
>
> I haven't been able to figure this out after spending 3 hours on
> Google.
>
> I can fetch an image, but I don't know where to put them.
> If I just use "new File("image.jpg")", it will end up in my tomcat
> bin/ folder.
>
> Some sugggests to use getServletContext().getRealPath("/"); but it
> would block forever at "getServletContext", I don't know why.
>
> What should I do so it will be part of the war, thus clients can
> access it via "http://path.to/appname/images/someimage.jpg";?
>
> What is the best practice on this?
>
> Thank you so much.
> Sam
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![http://path.to/appname/images/someimage.jpg"](http://path.to/appname/images/someimage.jpg"){kind=link}