Bootstrapping Spring Boot with a One-Liner (and a Tiny Python Script)

 Most Java developers don’t think twice about project creation.

You open Spring Initializr, click a few checkboxes, download a ZIP, extract it, and move on. It takes maybe a minute. Hardly worth automating — or is it?

This post is about a small, slightly unconventional tool: a Python script that hits Spring Initializr’s API, downloads a Spring Boot Maven project, and unzips it into a ready-to-use folder.

It’s not something you’ll use every day.
But when you do need it, it’s surprisingly handy.

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Why Would a Java Dev Care?

Let’s be honest:
Most of the time, the Spring Initializr website is perfectly fine.

However, there are a few situations where automation makes sense:

• You frequently create throwaway POCs

• You’re spinning up multiple microservices with similar configs

• You work on restricted or headless environments

• You want repeatable project templates

• You just enjoy tooling and automation (we all do)

In those cases, clicking through a UI starts to feel… unnecessary.

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Spring Initializr Is Already an API

What many developers don’t realize is that Spring Initializr is just an HTTP service.

That means you can generate a project with a single request:

curl https://start.spring.io/starter.zip \
  -d type=maven-project \
  -d dependencies=web,data-jpa \
  -d artifactId=demo \
  -o demo.zip

That’s it. No browser. No UI. No mouse.

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Taking It One Step Further with Python

The missing piece is unzip + folder setup.
That’s where a tiny Python script comes in.

This script:

1. Calls Spring Initializr

2. Downloads the ZIP

3. Extracts it into a clean project directory

4. Removes the ZIP file

All in one run.

import os
import zipfile
import urllib.request

PROJECT_NAME = "todo-app"
GROUP_ID = "com.todo"
PACKAGE_NAME = "com.todo.todoapp"
DEPENDENCIES = "web,data-jpa,mysql"

url = (
    "https://start.spring.io/starter.zip?"
    f"type=maven-project"
    f"&groupId={GROUP_ID}"
    f"&artifactId={PROJECT_NAME}"
    f"&name={PROJECT_NAME}"
    f"&packageName={PACKAGE_NAME}"
    f"&dependencies={DEPENDENCIES}"
)

zip_file = f"{PROJECT_NAME}.zip"

urllib.request.urlretrieve(url, zip_file)

with zipfile.ZipFile(zip_file, 'r') as zip_ref:
    zip_ref.extractall(PROJECT_NAME)

os.remove(zip_file)

print("Spring Boot project ready.")

Run it once, and you have a fully initialized Spring Boot Maven project — ready to open in your IDE.

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Is This Overkill?

For many developers: yes.

And that’s okay.

This isn’t meant to replace Spring Initializr’s UI.
It’s meant for those moments when you want:

• Consistency

• Speed

• Zero manual steps

• Scriptable project creation

It’s a tool you might forget about for months — and then suddenly appreciate when you need it.

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Where This Actually Shines

This approach works well when:

• You’re generating multiple services with the same baseline

• You want to bake project creation into automation scripts

• You’re teaching or demoing Spring Boot setup repeatedly

• You prefer infrastructure-as-code, even for small things

In short: it’s not flashy, but it’s practical.

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Final Thoughts

Good tools don’t have to be used all the time to be valuable.

This Python + Spring Initializr combo is one of those things:

• Small

• Simple

• Quietly useful

If you’re a Java developer who enjoys shaving off repetitive steps — or just likes knowing what’s possible under the hood — it’s worth keeping in your toolbox.

Even if you only use it once in a while.

Happy coding ☕