How to not minify a single variable?

Question

In my case, using AngularJS, i have a json with some configurations that can be used in controllers. The problem happens when i minify the application and the controller variables name does not match with the json.

Is there a way to not minify a single variabe?

My code: setting.json

{
   "settings": {
      "adIndex": 3
    }
}

controller:

function settingsService($q, $http, $log) {
  var settings = {
    adIndex: 0
  };

  var service = {
    init: init,
    getAdIndex: getAdIndex
  };
  return service;

  function init() {
    var deferred = $q.defer();

    $http.get('assets/settings/settings.json').then(function(response) {
      var loaded = response.data.settings;
      if (loaded) {
        settings = angular.extend(settings, loaded);
      }
      deferred.resolve(settings);
    }, function(error) {
      $log.info('Error loading settings', error);
      deferred.reject(error);
    });

    return deferred.promise;
  }

  function getAdIndex() {
    return settings.adIndex;
  }
}

UPDATE

After your suggestions i made some changes on code, but the problem not yet resolved. The code in controller is now:

function settingsService($q, $http, $log) {
  var settings = {
    'adIndex': 0
  };

  var service = {
    init: init,
    getAdIndex: getAdIndex
  };
  return service;

  function init() {
    var deferred = $q.defer();

    $http.get('assets/settings/settings.json').then(function(response) {
      var loaded = response.data['settings'];
      if (loaded) {
        settings = angular.extend(settings, loaded);
      }
      deferred.resolve(settings);
    }, function(error) {
      $log.info('Error loading settings', error);
      deferred.reject(error);
    });

    return deferred.promise;
  }

  function getAdIndex() {
    $log.debug('Object: ', settings); // result: settings { adIndex: 3 }
    $log.debug('Value: ', settings['adIndex']); // result: 0 
    return settings['adIndex'];
  }
}

How can 'adIndex' have 2 different values in logs?

That all depends on how your minifying the code? What minifyer are you using? — Liam, Mar 29 '17 at 15:47
You're trying to match hardcoded variable names to JSON returns? That sounds fragile... — Alexander Nied, Mar 29 '17 at 15:49
use strings, eg `settings['adIndex']` it won't modify that one — Icepickle, Mar 29 '17 at 15:50
It's not clear what's the problem here. What name doesn't match after minification? — JLRishe, Mar 29 '17 at 15:54
Side note: you're using the [explicit promise creation antipattern](http://stackoverflow.com/questions/23803743/what-is-the-explicit-promise-construction-antipattern-and-how-do-i-avoid-it) here. — JLRishe, Mar 29 '17 at 15:57

score 0 · Answer 1 · answered Mar 29 '17 at 16:01

0

  var settings = {
    "adIndex": 0  // quotes around the key, JSON style
  };

Not 100% sure how your minifier will handle this, but usually they can not modify string values. If the above doesn't work, you can do something slightly more verbose:

var settings = {};
settings['adIndex'] = 0;

In this case, the variable settings will be minified, but the string 'adIndex' will remain unchanged.

answered Mar 29 '17 at 16:01

SteamDev

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Thank for the answer, but not solve it. Question updated! – kilkus Mar 29 '17 at 17:00

How to not minify a single variable?

1 Answers1