AngularJS - 过滤searchQueries（复杂JSON）

Question

我实际上正在为我正在开发的WebApp做一个过滤器，问题是我无法设法为searchQuery设置一个有效的条件。

我实际上得到了那些东西：

myApp.filter('searchFilter', function() {
  return function(arrayToFilter, searchQuery, searchValue) {
    var baseArray = arrayToFilter;
    var usersSearchAfterQuery = [];
    var _searchQueryEmpty = {
      searchValue: searchValue,
      program: '',
      location: '',
      skillLevel: {
        shouldBeBeginner: false,
        shouldBeIntermediate: false,
        shouldBeExpert: false
      }
    };

    /**
     * true if the user has skillLevel at beginner
     * @param user
     * @returns {boolean}
     */
    function isBeginner(user) {
      return (user.skillLevel.toLowerCase() === 'beginner');
    }

    /**
     * true if the user has skillLevel at intermediate
     * @param user
     * @returns {boolean}
     */
    function isIntermediate(user) {
      return (user.skillLevel.toLowerCase() === 'intermediate');
    }

    /**
     * true if the user has skillLevel at expert
     * @param user
     * @returns {boolean}
     */
    function isExpert(user) {
      return (user.skillLevel.toLowerCase() === 'expert');
    }

    /**
     * true if object1 and object2 has same program
     * @param object1
     * @param object2
     * @returns {boolean}
     */
    function hasSameProgram(object1, object2) {
      return ((object1.program === object2.program));
    }

    /**
     * true if object1 and object2 has same location
     * @param object1
     * @param object2
     * @returns {boolean}
     */
    function hasSameLocation(object1, object2) {
      return ((object1.location === object2.location));
    }

    /**
     * returns true if query1 === query2
     * @param query1
     * @param query2
     * @returns {boolean}
     */
    function queriesAreEquals(query1, query2) {
      return (hasSameProgram(query1, query2)) &&     (hasSameLocation(query1, query2)) && (query1.skillLevel.shouldBeBeginner     === query2.skillLevel.shouldBeBeginner) &&     (query1.skillLevel.shouldBeIntermediate ===     query2.skillLevel.shouldBeIntermediate) &&     (query1.skillLevel.shouldBeExpert === query2.skillLevel.shouldBeExpert);
    }

    /**
     * if the filter is not empty or hasn't been reset
     */
    if (!queriesAreEquals(searchQuery, _searchQueryEmpty)) {
      console.log('Query is not empty');

      /**
       * second filter : if _user has the same values as in the filter
       */
      angular.forEach(arrayToFilter, function(_user) {

        if ((
              searchQuery.skillLevel.shouldBeBeginner && isBeginner(_user)) 
              || (searchQuery.skillLevel.shouldBeIntermediate && isIntermediate(_user)) 
              || (searchQuery.skillLevel.shouldBeExpert && isExpert(_user)
            ) 
            && (
                  hasSameLocation(_user, searchQuery) 
                  && hasSameProgram(_user, searchQuery)
                )
            ) {
              usersSearchAfterQuery.push(_user);
        }
      });
    }

    else if (searchValue !== null && typeof searchValue !== 'undefined' && searchValue.replace(/ /g, '') !== '') {
      if (typeof arrayToFilter !== 'undefined' && (searchValue.replace(/ /g, '') !== '' || searchValue.length !== 0)) {

        /**
         * first filter : if _user has the same skill as in searchValue
         */
        angular.forEach(arrayToFilter, function (_user) {
          if (_user.skill.toLowerCase().indexOf(searchValue.toLowerCase()) > -1) {
            usersSearchAfterQuery.push(_user);
          }
        });
      }
    } else {
      usersSearchAfterQuery = baseArray;
    }

    return usersSearchAfterQuery;
  };
});

因此，我必须将收到的searchQuery作为过滤器的参数进行比较，然后过滤arrayToFilter，之后视图应该只显示与查询匹配的用户。 searchValue是在搜索输入中输入的文本。

我的问题是：当我开始过滤查询时，我什么也没有显示。

我不知道自己做错了什么，或者我是否遇到了病情（我不希望这样，因为愚蠢的情况使我失去了4个小时）。

你们有些魔法药水吗？

非常感谢！

干杯。