Alert setup for container health in Docker - Mini Project: Build & Apply

Alert Setup for Container Health

📖 Scenario: You are managing a Docker container running a simple web application. You want to make sure the container is healthy and get alerted if it becomes unhealthy. This helps keep your app running smoothly.

🎯 Goal: Set up a Docker container with a health check and create a simple alert command to notify if the container is unhealthy.

📋 What You'll Learn

Create a Docker container running the nginx image

Add a health check command to the container configuration

Write a command to check the container's health status

Print an alert message if the container is unhealthy

💡 Why This Matters

🌍 Real World

Monitoring container health is important to keep applications running smoothly and avoid downtime.

💼 Career

DevOps engineers often set up health checks and alerts to maintain reliable containerized applications.

1

Create a Docker container running nginx

Write the docker run command to start a container named mynginx using the nginx image in detached mode.

Docker

# Your code here

Need a hint?

Use docker run -d --name mynginx nginx to start the container in the background.

2

Add a health check to the container

Write a docker run command to start a container named mynginx with a health check that runs curl -f http://localhost/ || exit 1 every 10 seconds, with a timeout of 5 seconds and retries set to 3.

Docker

docker run -d --name mynginx \
# Add health check options below
# Your code here

Need a hint?

Use --health-cmd, --health-interval, --health-timeout, and --health-retries options with docker run.

3

Check the container's health status

Write a docker inspect command to get the health status of the container named mynginx. Use jq to extract the health status from the JSON output.

Docker

docker run -d --name mynginx \
--health-cmd="curl -f http://localhost/ || exit 1" \
--health-interval=10s \
--health-timeout=5s \
--health-retries=3 \
nginx

# Your code here to check health status

Need a hint?

Use docker inspect --format='{{json .State.Health.Status}}' mynginx and save it to a variable.

4

Print an alert if the container is unhealthy

Write a bash if statement that checks if the variable health_status equals unhealthy. If yes, print "Alert: Container mynginx is unhealthy!".

Docker

docker run -d --name mynginx \
--health-cmd="curl -f http://localhost/ || exit 1" \
--health-interval=10s \
--health-timeout=5s \
--health-retries=3 \
nginx

health_status=$(docker inspect --format='{{json .State.Health.Status}}' mynginx | tr -d '"')

# Your code here to print alert if unhealthy

Need a hint?

Use if [ "$health_status" = "unhealthy" ] and echo to print the alert.