使用 forEach 时拼接不删除？答案

【问题标题】：splice is not deleting when using forEach?使用 forEach 时拼接不删除？
【发布时间】：2018-05-09 08:44:26
【问题描述】：

当产品 Line.Id 与退款 ID 匹配时，它应该从退款变量中删除该对象。

refund 变量应该只剩下 1 行，即：

refunds = {
  Lines: [
  {LineId: "444"}
  ]
}

我做错了什么？

查看示例：https://jsfiddle.net/ua8f1a5x/8/

 products = {
      Items: [{
          Name: "Item Name 1",
          LineId: "111",
          Status: [],
        },
        {
          Name: "Item Name 2",
          LineId: "222",
          Status: [],
        },
        {
          Name: "Item Name 3",
          LineId: "333",
          Status: [],
        }
      ]
    }
    
    refunds = {
      Lines: [
      {LineId: "222"},
      {LineId: "111"},
      {LineId: "444"}
      ]
    }
    
    
     refunds.Lines.forEach((refundItem, refundIndex) => {
        console.log("Checking Id " + refundItem.LineId);
         
         products.Items.forEach((Item) => {
             if (refundItem.LineId == Item.LineId) {
                 Item.Status.push({Name: "Refunded"});
                 
                 //Delete  object from refund
                 refunds.Lines.splice(refundIndex, 1);
             }
         });
     });
     
     
     console.log(refunds);
     console.log(products);

【问题讨论】：

请检查这个问题我认为它可能与Splicing a Javascript array from within the callback passed to forEach重复
在迭代数组时修改/删除数组中的项不是一个好主意
@GeorgeBailey 回答解决方案。

标签： javascript node.js

【解决方案1】：

forEach 迭代 static 数组，就像调用 forEach 时一样（并且如前所述，在迭代对象时对其进行变异会导致混淆）。在这种情况下，您应该改用filter：

const products = {
  Items: [{
      Name: "Item Name 1",
      LineId: "111",
      Status: [],
    },
    {
      Name: "Item Name 2",
      LineId: "222",
      Status: [],
    },
    {
      Name: "Item Name 3",
      LineId: "333",
      Status: [],
    }
  ]
}

const refunds = {
  Lines: [{
      LineId: "222"
    },
    {
      LineId: "111"
    },
    {
      LineId: "444"
    }
  ]
}

refunds.Lines = refunds.Lines.filter(refundItem => {
  console.log("Checking Id " + refundItem.LineId);
  const foundProduct = products.Items.find(({
    LineId
  }) => LineId === refundItem.LineId);
  if (foundProduct) {
    foundProduct.Status.push({
      Name: "Refunded"
    });
    return false;
  }
  return true;
});


console.log(refunds);
console.log(products);

【讨论】：

【解决方案2】：

你通常只使用 filter()

products = {
  Items: [{
      Name: "Item Name 1",
      LineId: "111",
      Status: [],
    },
    {
      Name: "Item Name 2",
      LineId: "222",
      Status: [],
    },
    {
      Name: "Item Name 3",
      LineId: "333",
      Status: [],
    }
  ]
}

refunds = {
  Lines: [{
      LineId: "222"
    },
    {
      LineId: "111"
    },
    {
      LineId: "444"
    }
  ]
}

refunds.Lines = refunds.Lines.filter(e => ! products.Items.some(x => x.LineId === e.LineId));
console.log(refunds)

【讨论】：

我的推送状态如何？

【解决方案3】：

我做错了什么？

refundIndex 在被迭代的数组也发生变异时不会被占用。

请改用filter。

refunds.Lines.filter( s => !products.Items.find( t => t.LineId == s.LineId ) );

演示

var products = {
  Items: [{
      Name: "Item Name 1",
      LineId: "111",
      Status: [],
    },
    {
      Name: "Item Name 2",
      LineId: "222",
      Status: [],
    },
    {
      Name: "Item Name 3",
      LineId: "333",
      Status: [],
    }
  ]
};

var refunds = {
  Lines: [
  {LineId: "222"},
  {LineId: "111"},
  {LineId: "444"}
  ]
};

refunds.Lines = refunds.Lines 
                .filter( s => !products.Items.
                    find( t => t.LineId == s.LineId ) );

console.log( refunds );

【讨论】：