Test a Weekly Cron Job

Test a weekly cron job

Just do what cron does, run the following as root:

run-parts -v /etc/cron.weekly

... or the next one if you receive the "Not a directory: -v" error:

run-parts /etc/cron.weekly -v

Option -v prints the script names before they are run.

How to run crontab job every week on Sunday

Here is an explanation of the crontab format.

# 1. Entry: Minute when the process will be started [0-60]
# 2. Entry: Hour when the process will be started [0-23]
# 3. Entry: Day of the month when the process will be started [1-28/29/30/31]
# 4. Entry: Month of the year when the process will be started [1-12]
# 5. Entry: Weekday when the process will be started [0-6] [0 is Sunday]
#
# all x min = */x

So according to this your 5 8 * * 0 would run 8:05 every Sunday.

How to unit test node-cron job

Here is the unit testing solution:

server.js:

const cron = require("node-cron");

const server = (module.exports = {
  cronJob: null,
  scheduledJob: function(pattern) {
    server.cronJob = cron.schedule(pattern, () => {
      server.run();
    });
  },
  run: function() {
    console.log("run called");
  },
});

server.test.js:

const server = require("./server");
const sinon = require("sinon");
const cron = require("node-cron");
const { expect } = require("chai");

describe("57208090", () => {
  afterEach(() => {
    sinon.restore();
  });
  describe("#scheduledJob", () => {
    it("should schedule job", () => {
      const pattern = "* * * * * *";
      const runStub = sinon.stub(server, "run");
      const scheduleStub = sinon
        .stub(cron, "schedule")
        .yields()
        .returns({});
      server.scheduledJob(pattern);
      sinon.assert.calledWith(scheduleStub, pattern, sinon.match.func);
      sinon.assert.calledOnce(runStub);
      expect(server.cronJob).to.be.eql({});
    });
  });

  describe("#run", () => {
    it("should run server", () => {
      const logSpy = sinon.spy(console, "log");
      server.run();
      sinon.assert.calledWith(logSpy, "run called");
    });
  });
});

Unit test result with 100% coverage:

  57208090
    #scheduledJob
      ✓ should schedule job
    #run
run called
      ✓ should run server


  2 passing (12ms)

----------------|----------|----------|----------|----------|-------------------|
File            |  % Stmts | % Branch |  % Funcs |  % Lines | Uncovered Line #s |
----------------|----------|----------|----------|----------|-------------------|
All files       |      100 |      100 |      100 |      100 |                   |
 server.js      |      100 |      100 |      100 |      100 |                   |
 server.test.js |      100 |      100 |      100 |      100 |                   |
----------------|----------|----------|----------|----------|-------------------|

You asked for unit testing. If you need an integration testing, please create a new post.

Source code: https://github.com/mrdulin/mocha-chai-sinon-codelab/tree/master/src/stackoverflow/57208090

Running aide --check as a crontab job once a week

A user id can only be specified in the system crontab file. The entries of a user's crontab file don't take a user id. The entries in question are apparently found in a user's crontab file, which is why you get root: command not found from the first, third and fourth entries.

From the second, you get cannot execute binary file because you ask bash to execute /usr/sbin/aide as a bash script when it's not a bash script. You should be using

*/1 * * * * /usr/sbin/aide --check

Crontab Day of the Week syntax

0 and 7 both stand for Sunday, you can use the one you want, so writing 0-6 or 1-7 has the same result.

Also, as suggested by @Henrik, it is possible to replace numbers by shortened name of days, such as MON, THU, etc:

0 - Sun      Sunday
1 - Mon      Monday
2 - Tue      Tuesday
3 - Wed      Wednesday
4 - Thu      Thursday
5 - Fri      Friday
6 - Sat      Saturday
7 - Sun      Sunday

Graphically, * * * * * command to be executed stands for:

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