


Hi,
How to list the content of a ZIP file with ZipFile?
In the documentation, we can only extract a file if we know its name (see ZipFile.getEntry(name) in https://docs.campaign.adobe.com/doc/AC/en/jsapi/c-ZipFile.html ).
How can we list the files inside? Or extract the first file? We tried ZipFile.entries, ZipFile.getEntries(), ZipFile[0] with no luck
Example:
var zip = new ZipFile(vars.filename);
logInfo(zip); // [object ZipFile]
logInfo(zip.open()); // function open() { [native code] }
logInfo(zip); // [object ZipFile]
logInfo(JSON.stringify(zip)); // {}
for(var prop in zip){
logInfo(prop, ':', zip[prop]); // fileCount : 1
}
logInfo(zip.getEntry('abc.zip')); // OK, but what if the file got an autogenerated timestapm
ZIP file:
At the moment we're using linux commands but this is somewhat a hack:
execCommand('unzip -j '+vars.filename+' -d target_directory');
Thank you
Florian
I've always just used execCommand(). Interesting doc link, that forms section looks like something new.
I've always just used execCommand(). Interesting doc link, that forms section looks like something new.