Javascript AngularJS-搜索查询过滤器（复杂JSON）

实际上，我正在为我正在开发的Web应用程序进行筛选，问题是我无法为searchQuery设置有效的条件

我真的得到了这些东西：

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，我必须比较作为过滤器参数接收的searchQuery，然后过滤arrayToFilter，然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，没有显示任何内容

我不知道我是否做错了什么，或者只是我的身体状况有问题（我希望没有，这愚蠢的身体状况折磨了我4个小时）

你们有魔药给我吗

非常感谢

干杯