zloBlog

I often sharing my files with use a file hosting. But I don't like use a browser for it. Now I've wrote the function for my .zshrc.

function zalil(){
        if [ -n "$1" -a -f "$1" ];then
                if  [ "`du -k $1|awk '{print $1}'`"  -lt "51200" ];then
                        curl -# -i -F "file"=@"$1" -F "submit=%20%20Send%20%20" http://zalil.ru/upload/|grep Location|sed 's/Location: /http:\/\/zalil.ru/'
                else
                        echo file $1 is large
                fi
        else
                echo file $1 is empty
        fi
}