Upload files to gofile.io via api. Required curl and jq

goupload

#!/bin/bash

#Simple test/help
if [[ "$#"  ==  '0' ]]
then
echo  -e '\nPlease Select File\n'

#Great, file selected.. Lets upload that..
elif [[ "$#"  ==  '1' ]]
then

#Find Best server to upload
srv=$(curl -s https://apiv2.gofile.io/getServer | jq  -r '.data|.server')

#Litlebit of hacking ;) (after upload, api respond with JSON that contain response code and file id)
upload=$(curl -X POST -# -F email=<MAIL_ADDRESS> -F description="${1%.*}"  -F filesUploaded=@"$1"  https://"$srv".gofile.io/upload | jq -r '.data|.code')

echo -e "\nYour Link: \nhttps://gofile.io/?c=$upload\n"

fi

you've forgotten the fiat the end

you've forgotten the fiat the end

yeah, exactly, it gives a syntax error
also sed can be used if you don't want to install jq

srv=$(curl -s https://apiv2.gofile.io/getServer | sed -E 's/.*?"server":"(.*?)".*?/\1/g')
upload=$(curl ... | sed -E 's/.*?"code":"(.*?)",.*?/\1/g')

and actually, only required parameter is file, if you dont want the other params,

srv=$(curl -s https://apiv2.gofile.io/getServer | sed -E 's/.*?"server":"(.*?)".*?/\1/g')
upload=$(curl -F [email protected] https://"$srv".gofile.io/upload | sed -E 's/.*?"code":"(.*?)",.*?/\1/g')

if you want to upload something quick without saving any scripts

curl -F [email protected] https://srv-file6.gofile.io/upload

link is gofile.io/d/code or gofile.io/?c=code