AngularJS Remove Substring from String in an expression {{}}

Question

I have this markup in my angularJS mark up

<div class="photo" style="background-image: url('{{item.mageURL}}')"></div>

The {{item.mageURL}} expression producing a string like:

http://example.com/photos/68678/stamp

I would like to change it to:

http://example.com/photos/68678

So far, without success, I've tried...

<div class="photo" style="background-image: url('{{item.mageURL.str.substring( 0, str.indexOf( 'stamp' ) )}}')"></div>

But it just returns an empty string.

Any idea how I can do this in AngularJS

I would suggest you to keep this kind of logic in your controller, and keep templates easier to read. Cheers. — Vitalii Strimbanu
– Vitalii Strimbanu, Commented Jan 11, 2017 at 19:32
I would suggest you not dirty up your controllers and instead either fix what your issue is (which it looks like you have a few solutions regarding the String APIs) OR utilize filters instead. — jusopi
– jusopi, Commented Jan 11, 2017 at 19:37

Mihai Alexandru-Ionut · Accepted Answer · 2017-01-11 19:39:43Z

2

You have to apply substring method to item.mageURL.str string.

'{{item.mageURL.str.substring( 0, item.mageURL.str.indexOf( '/stamp' ) )}}'

Another method is to use split method.

<div class="photo" style="background-image: url('{{item.mageURL.str.split('/stamp')[0]}}')"></div>

function MyCtrl($scope) {
  $scope.item={
      "mageURL":{
            "str":'http://example.com/photos/68678/stamp'
        }
  };
}

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<div ng-app>
  <div ng-controller="MyCtrl">
    <div class="photo" style="background-image: url('{{item.mageURL}}')"></div>
    {{item.mageURL.str.split('/stamp')[0]}}<br>
    {{item.mageURL.str.substring( 0, item.mageURL.str.indexOf( '/stamp' ) )}}
  </div>
</div>

edited Jan 11, 2017 at 19:39

answered Jan 11, 2017 at 19:30

Mihai Alexandru-Ionut

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2 Comments

jusopi Over a year ago

I like this best, if it works. Keep your controllers simple and UX driven. Keep logic like this either in your templates or use a filter.

Holly Over a year ago

ahh, that makes sense. Damn copy & past with str

Adrian Bratu · Accepted Answer · 2017-01-11 19:37:26Z

1

please use 'ng-style' https://docs.angularjs.org/api/ng/directive/ngStyle

ng-style="{'background-image' : 'url(' + item.mageURL.str.substring( 0, item.mageURL.str.indexOf( 'stamp' ) )+')'}"

<div class="photo" ng-style="{'background-image' : 'url(' +  item.mageURL.str.substring( 0, item.mageURL.str.indexOf( 'stamp' ) )+')'}"></div>

edited Jan 11, 2017 at 19:37

answered Jan 11, 2017 at 19:31

Adrian Bratu

5083 silver badges9 bronze badges

4 Comments

Holly Over a year ago

why is ng-style better in this case?

Adrian Bratu Over a year ago

in general performance is better with ng-style as it adds a watch on the model and will update the view only if the model has changed, in this case item.mageurl.str. When you use style it will execute everytime.

Holly Over a year ago

interesting, it is wrapped in a ng-repeat so I think it has a watch on it already?

Adrian Bratu Over a year ago

the ng-repeat watch triggers when count of the array / properties changes. ng-style has its own watch.

georgiy.zhuravlev · Accepted Answer · 2017-01-11 20:20:09Z

0

I think it'll be much better to create a tiny directive for this purpose, like this

app.directive('bgImg', function() {
  return function(scope, elem, attrs) {
    elem.css('background-image', someTransformFn(attrs.bgImg));
  }
})

Long JS-expressions in the templates are the real pain, in comparison with this:

<div bg-img="{{vm.someImg}}">

answered Jan 11, 2017 at 20:20

georgiy.zhuravlev

4857 silver badges23 bronze badges

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