Question

我使用http://fusejs.io/搜索一些json数据。这些数据是柜子的大清单，每个柜子都包含n个项目。

数据示例：

{"cabinets":[
  {"cabinet":"A4","items":[ 
    {"title":"Toys","keywords":"leksaker, pingpong, kendama, pingis","id":"98"},
    {"title":"Skisch Grand Prix","keywords":"Pokal, cup, pris, skisch","id":"99"},{"title":"Lamp prototype","keywords":"lamp-prototyp","id":"100"}]},
  {"cabinet":"B4","items":[
    {"title":"Screens (LCD)","keywords":"skärm display","id":"30"},
    {"title":"Displays","keywords":"screen skärm","id":"31"},
    {"title":"Lampor","keywords":"","id":"32"}]}
...

搜索会查看项目的标题和关键字，我希望它返回匹配项目的ID。

.then(function (response) {
  $scope.cabinets = response.data.cabinets;
  var options = {
    shouldSort: false,
    threshold: 0.3,
    location: 0,
    distance: 100,
    maxPatternLength: 32,
    minMatchCharLength: 2,
    keys: ["items.title","items.keywords"],
    id: "items.id"
  };

  fuse = new Fuse($scope.cabinets, options);
});

问题在于，当我搜索“显示”时，它会返回“屏幕（LCD”，这是机柜的第一项）的ID。如果我搜索“Lampor”，也会发生相同的事情。

根据http://fusejs.io/#searching-by-id

，这不是指定的行为

有人可以发现我的错误吗？

具有嵌套数据的Fuse.js返回错误的ID

0 个答案: