按多個屬性值過濾陣列

我有一個表單陣列

var cars = [
   {name: "BMW X5", topsales: ["USA", "China", "Russia"], maxspeed: 250, users: ["teenage", "ladies", "mens"]}
   {name: "Volkswagen Touareg", topsales: ["USA", "Germany"], maxspeed: 240, users: ["teenage", "mens", "old mens"]}
   etc....
]

我正在嘗試過濾，讓我們這樣說：

var query = {
   topsales: ["USA", "China"],
   users: "teenage"
}
function nestedFilter(targetArray, filters) {
     var filterKeys = Object.keys(filters);
     return targetArray.filter(function (eachObj) {
         return filterKeys.every(function (eachKey) {
             if (!filters[eachKey].length) {
                 return true; 
             }
             return filters[eachKey].includes(eachObj[eachKey]);
          });
     });
};
goodresult = nestedFilter(cars, query);

但該功能無法正常作業。如果物件在屬性中有一個值，則它進行過濾，但如果有多個值，并且我至少需要其中一個來滿足搜索，則它不進行過濾。幫助誰可以請

uj5u.com熱心網友回復：

您可以檢查查詢是否為陣列和/或值是否為陣列并進行相應檢查。

function nestedFilter(data, query) {
    const
        filters = Object.entries(query);

    return data.filter(o => filters.every(([k, v]) => Array.isArray(v)
        ? Array.isArray(o[k])
            ? v.some(s => o[k].includes(s))
            : v.includes(o[k])
        : Array.isArray(o[k])
            ? o[k].includes(v)
            : o[k] === v
    ));
}

const
    cars = [{ name: "BMW X5", topsales: ["USA", "China", "Russia"], maxspeed: 250, users: ["teenage", "ladies", "mens"] }, { name: "Volkswagen Touareg", topsales: ["USA", "Germany"], maxspeed: 240, users: ["teenage", "mens", "old mens"] }],
    query = { topsales: ["USA", "China"], users: "teenage" };

console.log(nestedFilter(cars, query));

uj5u.com熱心網友回復：

我假設您打算實作一項OR功能，因為您至少說了其中之一。因此，作業代碼如下。

但在繼續閱讀之前，請注意以下備注：

• 我使用了some而不是every，因為some作業方式or和 every作業方式and。這意味著如果當前car專案匹配至少一個過濾器，該行將回傳 true 。

• 您應該使用item.includes(filter)而不是filter.includes(item).

• 您需要檢查當前過濾器項是否為陣列，并采取相應措施。

• 在這段代碼中，我沒有處理它并假設它currentCandidate是一個字串或一個基元。如果在其他情況下候選專案（即 的欄位car）本身也是一個陣列，那么您必須更新代碼來處理它。

  var cars = [
    {name: "BMW X5", topsales: "USA, China, Russia", maxspeed: 250, users: "teenage, ladies, men"},
    {name: "Volkswagen Touareg", topsales: "USA, Germany", maxspeed: 240, users: "teenage, men, old men"}
  ]
  
  var query = {
    topsales: ["USA", "China"],
    maxspeed: 240
  }
  
  function nestedFilter(targetArray, filters) {
    const filterKeys = Object.keys(filters);
    return targetArray.filter(function (eachObj) {
      //using some instead of every to make sure that it works as OR
      const result = filterKeys.some(function (eachKey) {
        //the current item that we are trying to use in the filter
        const currentCandidate = eachObj[eachKey];
  
        //the current item that we are using as a filter
        const currentFilterItem = filters[eachKey]
  
        if (Array.isArray(currentFilterItem)) {
          if (currentFilterItem.length === 0) {
            //no filter, return true
            return true
          }
  
          //loop on each item in the currentFilterItem
          //if any of them matches simply return true (OR)
          for (let filterKey in currentFilterItem) {
            if (currentCandidate.includes(currentFilterItem[filterKey])) {
              return true
            }
          }
          //for loop ended, no match
          return false
        } else {
          //the current filter item is not an array, use it as one item
          //return eachObj[eachKey].includes(currentFilterItem)
          return currentCandidate === currentFilterItem
        }
      });
  
      return result;
    });
  }
  
  goodresult = nestedFilter(cars, query);
  console.debug(goodresult)

uj5u.com熱心網友回復：

您可以檢查“filterKey”的值是否不是陣列，將其設為陣列，并檢查陣列是否有子陣列

function hasSubArray(master, sub) {
  return sub.every((i => v => i = master.indexOf(v, i)   1)(0));
}

function nestedFilter(targetArray, filters) {
  var filterKeys = Object.keys(filters);
  return targetArray.filter(function (eachObj) {
    return filterKeys.every(function (eachKey) {
      var subArray = filters[eachKey];
      if (!Array.isArray(filters[eachKey])) {
        subArray = [filters[eachKey]];
      }

      return hasSubArray(eachObj[eachKey], subArray);
    });
  });
}

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