angular.js - ng-if defines a method and returns a string, why is it wrong?

Question

This is the right thing to write

<li ng-repeat="(x, y) in item" ng-if="x!='a'"></li>

Writing like this is to output all, ng-if is not executed

<li ng-repeat="(x, y) in item" ng-if="test()"></li>

...

$scope.test = function(){
    return "x!='a'"
}

Answer 1

Firstly, you did not pass x in, and secondly, what you returned was a string, and the string is always correct. Try writing it like this:

<li ng-repeat="(x, y) in item" ng-if="test(x)"></li>

$scope.test = function(x){
    return x!='a';
}

Answer 2

Tell you a concept, the things in ng-if are called angular expressions, and angular will parse this expression.
"x!='a'" is actually $scope.x != 'a'. By the way, this $scope is ng-repeat. "x!='a'"其实就是$scope.x != 'a'，顺带一提这个$scope是ng-repeat产生的scope。
下面的test()当然会变parse成$scope.test()在ngRepeat的scope中没找到方法，所以从父scope中找到了你的方法， 然后你的方法return的是一个字符串， 所以判断永远是trueThe following test() will of course be parsed into $scope.test(). No method was found in the scope of ngRepeat, so you found your method from the parent scope. , then your method returns a string, so the judgment is always true.