角度筛选列表对象到其他列表

Question

我有一件定制的物品：

export class PartsChildInfo {
   name: string;
   materialName: string;
   thickNess: number;
}

export class PartGroupInfo
{
   materialName: string;
   thickNess: number;
}

例如，我有一个列表项PartsChildInfo

list : PartsChildInfo  = [
  { Name = "GA8-0608" , MaterialName = "SS"  , ThickNess = 1 };
  { Name = "05F1-051" , MaterialName = "SUS" , ThickNess = 2 };
  { Name = "2B73-002" , MaterialName = "AL"  , ThickNess = 3 };
  { Name = "01-20155" , MaterialName = "SS"  , ThickNess = 1 };
  { Name = "02MEG099" , MaterialName = "SUS" , ThickNess = 2 }; 
]

我希望在MaterialName中获得如下列表，ThickNess与list相同：

testChildList : PartGroupInfo = [
  { MaterialName = "SS"  , ThickNess = 1 };
  { MaterialName = "SUS" , ThickNess = 2 };
  { MaterialName = "AL"  , ThickNess = 3 }; 
]

我试过这个

testChildList : PartGroupInfo[] = [];
for (let i = 0; i < list.length; i++) {
   let targeti = list[i];
   for (let j = 0; j < this.testChildList.length; j++) {
      let targetj = this.testChildList[j];
      if (targeti.materialName != targetj.materialName && targeti.thickNess != targetj.thickNess) {
        let item = new PartGroupInfo();
        item.materialName = targeti.materialName;
        item.thickNess = targeti.thickNess;
        this.testChildList.push(item);
       }
    }
 }

但返回的列表为空。我该怎么修呢？

Answer 1

也许使用.forEach来迭代list数组，通过index检查条目是否存在于testChildList中。当索引为-1 (不存在)时，将item推到testChildList。

this.list.forEach((item) => {
  var i = this.testChildList.findIndex(
    (x) =>
      x.materialName == item.materialName && x.thickNess == item.thickNess
  );

  if (i == -1)
    this.testChildList.push({
      materialName: item.materialName,
      thickNess: item.thickNess,
    });
});

基于StackBlitz的样本解决方案