Revolutionize Web Development with HTMX : Building a Multi-Upload File Service with Golang Echo (HTMX Edition)

3 min readMar 14, 2024

Introduction

In this comprehensive guide, we’ll explore the fusion of Golang, Echo framework, and htmx to create a dynamic Multi-Upload File Service. Whether you’re a seasoned coder or just getting started, this tutorial will equip you with the knowledge and tools to build robust web applications with ease. Let’s dive in and harness the power of htmx together!

Prerequisites

Before we begin, ensure you have Golang installed on your system. Familiarity with basic web development concepts such as HTML, CSS, and JavaScript will be beneficial but not required.

Setting Up Your Project

Let’s kick things off by setting up a new Golang project and installing the necessary dependencies. Follow these steps in your terminal:

mkdir file-upload-service-htmx
cd file-upload-service-htmx
go mod init file-upload-service-htmx
go get -u github.com/labstack/echo/v4

We’re leveraging Go modules for dependency management and installing Echo version 4.

Creating the File Upload Service

Next, let’s create the main file for our file upload service. Create a file named main.go in your project directory and add the following code:

package main

import (
    "io"
    "net/http"
    "os"

    "github.com/labstack/echo/v4"
)

func uploadFile(c echo.Context) error {
    // Read form file
    file, err := c.FormFile("file")
    if err != nil {
        return err
    }
    // Source
    src, err := file.Open()
    if err != nil {
        return err
    }
    defer src.Close()

    // Destination
    dst, err := os.Create(file.Filename)
    if err != nil {
        return err
    }
    defer dst.Close()

    // Copy
    if _, err = io.Copy(dst, src); err != nil {
        return err
    }

    return c.String(http.StatusOK, "File uploaded successfully")
}

func main() {
    e := echo.New()

    // Routes
    e.POST("/upload", uploadFile)
    e.GET("/upload-status", func(c echo.Context) error {
        // Dummy endpoint for upload status
        return c.String(http.StatusOK, "Upload status: Success")
    })

    // Start server
    e.Start(":8080")
}

In this code snippet:

We import the necessary packages, including Echo.
The uploadFile function handles file uploads, similar to the previous version.
We define a simple main function where we initialize our Echo instance, define routes for file upload and upload status, and start the server on port 8080.

Enhancing the User Interface with htmx : Now, let’s take our file upload form to the next level using htmx for dynamic behavior. Create a file named upload.html in your project directory and add the following code:

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>File Upload</title>
    <script src="https://unpkg.com/htmx.org@1.7.0/dist/htmx.min.js"></script>
</head>
<body>
    <h1>File Upload</h1>
    <form hx-post="/upload" hx-target="#upload-status" enctype="multipart/form-data">
        <input type="file" name="file" id="file">
        <button type="submit" hx-post="submit">Upload</button>
    </form>
    <div id="upload-status" hx-get="/upload-status"></div>
</body>
</html>

In this HTML code:

We include the htmx library from a CDN.
We’ve removed the action attribute from the form tag since htmx handles form submission through AJAX.
We use hx-post attribute on the form tag to specify the endpoint for file upload.
We’ve added an empty div with the id upload-status where we'll display the upload status.
The hx-target attribute specifies where the response of the form submission will be inserted.

Conclusion

Congratulations! You’ve successfully created a dynamic Multi-Upload File Service using Golang Echo and htmx. With htmx, you can add interactivity to your web applications without the need for complex JavaScript code. Keep exploring the endless possibilities and stay tuned for more innovative solutions in the world of web development!