Alon's questions

The following bash command:

aws s3 ls s3://bucket/dir | grep _SUCCESS | wc -l

prints:

0

What I want is, when the result is indeed 0, to print FAILURE to the terminal.

This is what I've tried:

if [ aws s3 ls s3://bucket/dir | grep _SUCCESS | wc -l = 0]; then echo FAILURE; fi

but nothing is printed. it looks like it is waiting for more input.

Directory structure:

/home
---ubuntu
------projects
---------project1
---------project2
---------project3
------.ignore

.ignore:

*project1* *project2*

If I run:

zip -r zipush.zip projects -x "$(< .ignore)"

Then it creates a zip containting all three projects - which is a wrong result. I want it to ignore project1 and project2.

However, it does work well when I write the arguments explicitly:

zip -r zipush.zip projects -x *project1* *project2*

But I don't understand why, because "$(< .ignore)" seems to work well when I type for example:

echo "$(< .ignore)"

Output:

 *project1* *project2*

I launched an Ubuntu 18 instance on EC2.

I connected to the server using ubuntu user.

I created a new user named newuser and added it to the sudo group.

Then I ran:

rsync --archive --chown=newuser:newuser ~/.ssh /home/newuser

in order to let the new user to connect to the server directly.

When I'm trying to connect to the server using the new user I'm getting the following error:

Using username "newuser".

Authenticating with public key "imported-openssh-key"

Please login as the user "ubuntu" rather than the user "root".

I get the same error from both MobaXterm and PuTTY.

I've downloaded uMongo to my Ubuntu 14 OS.

I've extracted the zip to a directory.

When I double click it, I just get a text editor opened.

When I try to open it from terminal, nothing is happening:

username:~/Software/umongo-linux-all_1-6-2$ chmod +x launch-umongo.sh
username:~/Software/umongo-linux-all_1-6-2$ 

Any help will be profoundly appreciated!