javascript - 这些重复的代码有没有办法简化下？js

Question

$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';
        });

Answer 1

    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))
})

Answer 2

$scope.result = {};

$scope.pls.forEach(function(item) {
  var idArr = item.id.split('-');

  var key = idArr[1].toLowerCase() + '_' + idArr[2].toLowerCase();
  if (!$scope.result[key]) {
    $scope.result[key] = [];
  } else {
    $scope.result[key].push(item);
  }
});

Answer 3

我这个也凑合用吧^^

const keys = ['EZZH', 'BSDW', 'SGDW', 'BGHS', 'SGHS'];

const values = ['QSW', 'ZSW', 'HSW', 'QSW'];


keys
    .forEach(k => {
        values.forEach(v => {
            $scope[`${k.toLowerCase()}_${v.toLowerCase()}`] = $scope.pls.filter(item => {
                return item.id.split('-')[1] === k && item.id.split('-')[2] === v;
            });
        });
    });

console.log($scope);

Answer 4

一千变量一千filter，一个for循环搞定的事儿

Answer 5

各位都是高手啊