javascript - 这些重复的代码有没有办法简化下？js

问题描述

问 题

$scope.ezzh_qsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'QSW';
        });
        $scope.ezzh_zsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'ZSW';
        });
        $scope.ezzh_hsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'HSW';
        });
        $scope.bsdw_qsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'QSW';
        });
        $scope.bsdw_zsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'ZSW';
        });
        $scope.bsdw_hsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'HSW';
        });
        $scope.bgdw_qsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'QSW';
        });
        $scope.bgdw_zsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'ZSW';
        });
        $scope.bgdw_hsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'HSW';
        });
        $scope.sgdw_qsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'QSW';
        });
        $scope.sgdw_zsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'ZSW';
        });
        $scope.sgdw_hsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'HSW';
        });
        $scope.bghs_qsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'QSW';
        });
        $scope.bghs_zsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'ZSW';
        });
        $scope.bghs_hsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'HSW';
        });
        $scope.sghs_qsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'QSW';
        });
        $scope.sghs_zsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'ZSW';
        });
        $scope.sghs_hsw = $scope.pls.filter(function(item, index) {
          return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == 'HSW';
        });

解决方案

    var filter = function(text){
        return function(item, index) {
              return item.id.split('-')[1] == 'EZZH' && item.id.split('-')[2] == text;
            }
    }
    
    // 使用
    $scope.sghs_hsw = $scope.pls.filter(filter("HSW"))

还有更简洁的. 结合上面的:

var map = {
    sghs_hsw:'HSW',
    sghs_qsw :'QSW',
    /// 省略其它的
}
_.each(map,function(v,k){
    $scope[k] = $scope.pls.filter(filter(v))
})

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