I did not test anything, but a quick look to the java.util.zip package learns me that there is a method getNextEntry in the ZipInputStream. You can of coure wrap the jcifs inputstream to a ZipInputStream and then loop the ZipEntries untill you are at the position of the file you want. If you're lucky, the file is not located at the end of the stream :)<div>
<div><br><div class="gmail_quote">On Tue, Jul 14, 2009 at 11:23, <a href="mailto:pedrofaundezgon@terra.es">pedrofaundezgon@terra.es</a> <span dir="ltr"><<a href="mailto:pedrofaundezgon@terra.es">pedrofaundezgon@terra.es</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">I want to access the contents of a Zip file from a Java program.<br>
I know java.util.zip.ZipFile can only access local files.<br>
And I also know the default API exposed by jcifs is only <a href="http://java.io" target="_blank">java.io</a>.<br>
InputStream API<br>
Is there an easy way to read a huge ZipFile over the network<br>
extracting only the file I am interested in the most efficient way<br>
possible?<br>
Is there a way to have jcifs expose a random access file interface<br>
(just like local file) and a way to have ZipFile consume it?<br>
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<br>
<br>
Ahora también puedes acceder a tu correo Terra desde el móvil.<br>
Infórmate pinchando aquí.<br>
<br>
<br>
</blockquote></div><br><br clear="all"><br>-- <br>--<br>Palanthir BV<br>Arjan van der Veen<br>Krepelsbosch 129<br>7325AP Apeldoorn<br>+31655730384<br><a href="mailto:avdveen@palanthir.nl">avdveen@palanthir.nl</a><br>
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