How can I replace minified JavaScript files with non-minified versions in AngularJS for local development purposes?

What is the most effective approach to tackle this issue with AngularJS?

I am currently working on an AngularJS application that consists of concatenated/minified JavaScript files. However, in my development environment, I prefer loading the non-concatenated/non-minified versions of these files for easier debugging purposes.

<!-- production -->
<script src="js/all.min.js"></script> <!-- contains all concatenated/minified code -->

<!-- dev -->
<script src="js/services.js"></script>
<script src="js/controllers.js"></script>
<script src="js/directives.js"></script>

I have been struggling with finding a suitable conditional mechanism in AngularJS to achieve this. One solution I considered was using a directive on each script tag.

Currently, I use Gulp to concatenate and uglify the files, but I have not been able to find a method within Gulp to accommodate this requirement.

It's worth noting that PHP is the server-side language being used.

Any assistance or guidance on this matter would be highly appreciated.

Answer №1

To enhance your debugging process, consider generating a "Source Map" in addition to the minified javascript file. By utilizing the source map, you can easily refer back to the original code while troubleshooting. Tools like UglifyJS 2.0 or Closure Compiler offer the option to create a source map alongside the minified javascript.

For further insights on source maps and debugging techniques: explore source maps debugging in Google Chrome

Answer №2

To effectively implement this in Angular, one approach might be to utilize ngInclude for conditionally loading an ngTemplate. The code snippet below demonstrates how this can be achieved:

<script type="text/ng-template" id="development">
  <script src="js/services.js"></script>
  <script src="js/controllers.js"></script>
  script src="js/directives.js"></script>
</script>

<script type="text/ng-template" id="production">
  <script src="js/all.min.js"></script>
</script>

<div id="javascripts" ng-include="dev ? 'development' : 'production'"></div>

If your controller involves the use of $scope, simply set $scope.dev to true for including development JavaScript, or false to opt for the production file.

It may also be advisable to explore task runners for a more efficient method. Additionally, consider leveraging environment variables to dictate the project's current state.

Incorporating Gulp for this purpose, you can make use of the gulp-html-replace package. A sample task utilizing this package is outlined below:

var gulp = require('gulp');
var htmlreplace = require('gulp-html-replace');

gulp.task('replace-js', function(){
if (process.env.NODE_ENV === 'production') {
    gulp.src('index.html')
      .pipe(htmlreplace({
        'js': ['js/services.js', 'js/controllers.js', 'js/directives.js']
      }))
  }
});

Given this HTML structure:

<!-- build:js -->
<script src="js/all.min.js"></script>
<!-- endbuild -->

(Please note that the following instructions involve node-specific operations and may need adjustment according to your system)

Upon running Gulp, it is imperative to set the environment variable, as seen in the if statement above with process.env.NODE_ENV. This can be done by executing Gulp as follows:

set NODE_ENV=production && gulp replace-js

Depending on your operating system, you may need to refer to specific guidelines like "set environment variable node YOUR_OS" in case any issues arise.

For easier access, add the command to your package.json under the scripts key, incorporating a setup similar to the example provided below:

"scripts": {
  "replace-js": "set NODE_ENV=production && gulp replace-js"
}

This way, you can execute the designated command effortlessly instead of manually typing out the entire expression:

npm run replace-js

Please bear in mind that these details are based on memory and assuming no errors have been made, ensuring clarity for seamless execution.

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