getEntry

Extracts a file from the archive.


            getEntry (filename)

Parameters

filename

Name of the file in the archive.

Return value

The ⌐ MemoryBuffer that contains the contents of the file.

Remarks

The archive must have been opened with the ⌐open() method before you can use this method. Otherwise, an exception is raised. Example: The following example extracts the dir/content.xml file from an archive and saves it to the /tmp/ folder.

var zip = new ZipFile("/tmp/archive.zip")
zip.open()

var content = zip.getEntry("dir/content.xml")
content.save("/tmp/content.xml")

zip.dispose()
content.dispose()

Features

Method of class: ZipFile

Available in:

Content management
Delivery properties
Typology rule
JSSP
SOAP Method
WebApp
Workflow