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:
Available in:
- Content management
- Delivery properties
- Typology rule
- JSSP
- SOAP Method
- WebApp
- Workflow