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In my controller I can call:
$scope.games[0];
To access the first item in my games array. Is there a way to do this keeping Filters in mind.
For example I have:
filter:search
On my repeat, how can I call $scope.list[0]; to equal the first search result?
This has been answered, thank you. I built a cool widget using AngluarJs for Lagged.com, play free online games here: https://lagged.com/
956
This can be done by injecting the filter's dependency into a controller and calling it in code like
var filteredArray = filterDependency(arrayToFilter,args);
which returns a new, filtered array. Since you are using the "filter" filter (it's a filter whose name is filter), the dependency injection should be filterFilter. Your controller should look something like this:
var app = angular.module('myapp',[]);
app.controller('ctrlParent',function($scope,filterFilter){
var filteredArray = [];
$scope.list = ["abc","def","ghi","abcdefghi"];
$scope.$watch('search',function(newValue){
filteredArray = filterFilter($scope.list, newValue);
// do something with the first result
console.log(filteredArray[0]); // first result
});
});
What we're doing is setting a watch on the input model (search) so we can get the new value and re-filter the array any time the input is changed.
Also:
If you need to access the ng-repeat index from within the view, you can use the special property $index inside of the ng-repeat like:
<div ng-repeat="item in list | filter:search">
{{$index}}
</div>
You can also use $first, $middle, and $last as shown in this Angular doc.
Demo: Here is a fiddle
88
Not with bracket access. If you have an ng-repeat with a filter:
ng-repeat="thing in things | filter:search"
The filtered list here is kind of anonymous - it doesn't have a name that you can access.
That said, if you take a look at the docs for ngRepeat, you'll see that inside each repeater's scope, you have access to $index, $first, $middle, and $last.
So something like
<body ng-app="App" ng-controller="Main">
<pre ng-repeat="n in nums | filter:isOdd">
n={{n}}:index={{$index}}:first={{$first}}:middle{{$middle}}:last={{$last}}
</pre>
</body>
Would yield:
n=1:index=0:first=true:middlefalse:last=false
n=3:index=1:first=false:middletrue:last=false
n=5:index=2:first=false:middlefalse:last=true
Fiddle
3
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