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Download all the resources available under public resource

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Level 6

Hi Team, 


Could all the data within the public resources, including subfolders and individual files, be exported using a query activity & Export activity? Alternatively, is it achievable through an API call? Your suggestions would be appreciated. Thank you in advance.

 

 

rvnth_0-1714729751266.pngrvnth_1-1714729789314.png

 


I'm observing the count, but how can I verify it when dealing with multiple subfolders? Additionally, is it feasible to retrieve the files along with their respective subfolders as column names?

Instead of downloading a CSV containing data about these files, I prefer to download all the files themselves.

1 Accepted Solution

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Correct answer by
Community Advisor

Hi @rvnth 

 

You can write Linux/Unix code to download the full directory. You can refer the link for more reference.

scp -r username@<public ip>:/<path to source file> /<path to destination>

 

Hope this helps.

 

Regards

Akshay

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3 Replies

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Community Advisor

Hi @rvnth ,

 

Could you Please let us know if this exercise is just to verify the Data, you can export to a .CSV File with the below mentioned column names in the mentioned Query's Additional data with below Columns:

 

LakshmiPravallika_0-1714747435743.png

This will give all the Information of Folders and Sub Folders and the Full URL generated after publishing the files.

 

You can check randomly for few records whether the "Unique File name" (URL) generated for the files is working properly in the browser.

 

Regards,

Pravallika.

 

 

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Level 6

Hi @LakshmiPravallika ,

 

I'm interested in downloading all the available image files rather than just the metadata. Could you please inform me if there's a method to accomplish this via API or by importing it as a package? Thank you in advance.

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Correct answer by
Community Advisor

Hi @rvnth 

 

You can write Linux/Unix code to download the full directory. You can refer the link for more reference.

scp -r username@<public ip>:/<path to source file> /<path to destination>

 

Hope this helps.

 

Regards

Akshay