Your achievements

Level 1

0% to

Level 2

Tip /
Sign in

Sign in to Community

to gain points, level up, and earn exciting badges like the new
Bedrock Mission!

Learn more

View all

Sign in to view all badges

SOLVED

List Zip file content with ZipFile

Avatar

Community Advisor

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:

2019-01-17_110455_screen_2.jpg

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

1 Accepted Solution

Avatar

Correct answer by
Community Advisor

I've always just used execCommand(). Interesting doc link, that forms section looks like something new.

View solution in original post

1 Reply

Avatar

Correct answer by
Community Advisor

I've always just used execCommand(). Interesting doc link, that forms section looks like something new.