Question

我有以下JSON

[  
   {  
      "Id":1068,
      "Name":"jone",
      "SortOrder":0
   },
   {  
      "Id":1077,
      "Name":"John ",
      "SortOrder":0
   },
   {  
      "Id":1074,
      "Name":"sasasa",
      "SortOrder":0
   },
   {  
      "Id":1078,
      "Name":"fdfdfd",
      "SortOrder":0
   },
   {  
      "Id":1029,
      "Name":"fdfdfd",
      "SortOrder":0
   },
   {  
      "Id":1076,
      "Name":"sdf",
      "SortOrder":0
   },
   {  
      "Id":1030,
      "Name":"erer",
      "SortOrder":0
   },
   {  
      "Id":11,
      "Name":"tryryr",
      "SortOrder":0
   },
   {  
      "Id":1008,
      "Name":"bvbf",
      "SortOrder":0
   },
   {  
      "Id":9,
      "Name":"jkk",
      "SortOrder":0
   },
   {  
      "Id":19,
      "Name":"pioku",
      "SortOrder":0
   },
   {  
      "Id":5,
      "Name":"cssxs",
      "SortOrder":0
   },
   {  
      "Id":1009,
      "Name":"mhmh",
      "SortOrder":0
   }
]

现在，我需要通过传递ID从JSON中删除商品。例如，当我向函数传递1068,1077 id（逗号分隔）时，我需要从JSON中删除相关数据。

[ { "Id":1074, "Name":"sasasa", "SortOrder":0 }, { "Id":1078, "Name":"fdfdfd", "SortOrder":0 }, { "Id":1029, "Name":"fdfdfd", "SortOrder":0 }, { "Id":1076, "Name":"sdf", "SortOrder":0 }, { "Id":1030, "Name":"erer", "SortOrder":0 }, { "Id":11, "Name":"tryryr", "SortOrder":0 }, { "Id":1008, "Name":"bvbf", "SortOrder":0 }, { "Id":9, "Name":"jkk", "SortOrder":0 }, { "Id":19, "Name":"pioku", "SortOrder":0 }, { "Id":5, "Name":"cssxs", "SortOrder":0 }, { "Id":1009, "Name":"mhmh", "SortOrder":0 } ]

当将1009,1068,1074,1078,1029,1076,1030,11,1008传递到JSON后面的函数时，将得到结果。

[  
   {  
      "Id":9,
      "Name":"jkk",
      "SortOrder":0
   },
   {  
      "Id":19,
      "Name":"pioku",
      "SortOrder":0
   },
   {  
      "Id":5,
      "Name":"cssxs",
      "SortOrder":0
   },
   {  
      "Id":1009,
      "Name":"mhmh",
      "SortOrder":0
   }
]

什么是最有效的方法？并请注意。我不能使用ES6或ES5，因为某些ES5，ES6功能无法在IE和Safari浏览器中使用。

Answer 1

这可以解决问题。可以使用诸如.map或.filter之类的Array方法使它变得不那么冗长。

var myFilter = [1009,1068,1074,1078,1029,1076,1030,11,1008];
var myData = [
  { "Id":1074, "Name":"sasasa", "SortOrder":0 },
  { "Id":9, "Name":"jkk", "SortOrder":0 },
  { "Id":19, "Name":"pioku", "SortOrder":0 },
  { "Id":5, "Name":"cssxs", "SortOrder":0 },
  { "Id":1009, "Name":"mhmh", "SortOrder":0 }
];
function filterOut(data, filter){
  result = [];
  for (let i = 0; i < data.length; i++){
    if(filter.indexOf(parseInt(data[i].Id)) == -1){
      result.push(data[i]);
    }
  }
  return(result);
}

var output = filterOut(myData, myFilter);
console.log(output);

（请注意，您期望的结果不正确，因为其中包含1009，这是您要求过滤掉的ID值之一。为简便起见，我从列表中删除了一些数据对象。）

编辑： 我刚刚注意到您的“没有现代JS”要求。我想我已经从ES5开始删除了所有内容。

Answer 2

我的建议是编写最简单的可行代码，然后进行polyfill或transpile以支持您的目标环境。

请记住，这是一个用现代风格编写的简单版本：

const removeByIds = (ids, idNbrs = ids.split(',').map(Number)) => 
  (data) => data.filter(({Id}) => !idNbrs.includes(Id))

const data = [{"Id": 1077, "Name": "John ", "SortOrder": 0}, {"Id": 9, "Name": "jkk", "SortOrder": 0}, {"Id": 19, "Name": "pioku", "SortOrder": 0}, {"Id": 5, "Name": "cssxs", "SortOrder": 0}]

console .log (
  removeByIds ('1009,1068,1074,1078,1029,1076,1030,11,1008') (data)
)

如何通过传递ID从JSON中删除项目

2 个答案: