How can I change crontab dynamically? [closed]

Question

I need to create a bash script in order to:

1. Empty my root Crontab;
2. Insert new Cronjobs via bash script.

For the first point I can use crontab -r

For the second point instead here I found this script:

#!/bin/bash

lines="* * * * * /path/to/command"
(crontab -u root -l; echo "$lines" ) | crontab -u root -

How can I cook this together in a bash script?

Something like this:

#!/bin/bash 

crontab -r 

line="* * * * * /path/to/command; 
      * * * * * /path/to/command2;
      * * * * * /path/to/command3" 

(crontab -u root -l; echo "$line" ) | crontab -u root -

Answer 1

The sample you posted would print current crontab and inject new directives.

If you intend to just inject new directives, wiping the current crontab, instead of your

lines="* * * * * /path/to/command"
( crontab -u root -l; echo "$lines" ) | crontab -u root -

Go with:

lines="* * * * * /path/to/command"
echo "$lines" | crontab -u root -

---

And, as you pointed it out in the comments, it is wrong, adding multiple crons, to use semicolons as a separator. You can go with:

lines=" line1
line2"

Or:

crontab -u root - <<EOF
line1
line2
EOF

Or:

(
    echo line1
    echo line2
) | crontab -u root -