无法验证所有嵌套子元素长度

Question

我有1个对象包含了像下面嵌套子：

$scope.artists.materials.items[] 

现在我有几个艺术家这将包含的项目列表，但在这我要检查总每个艺术家项目的长度，如果发现不匹配，那么我想返回true或false。

Problem is when i dont have items for any of the artist then still i am getting false

这里的想法是存储来自第一位艺术家的物品的长度，并确保他们都具有相同的物品长度。

代码：

function checkItemsValidity() { 
     for (var i = 1; i < $scope.artists.length; index++) { 
      if ($scope.artists[i].materials.items != undefined && $scope.artists[0].materials.items) { 
       if($scope.artists[i].materials.items.length != $scope.artists[0].materials.items[0].length) { 
          return false; 
       } 
      }   
          return false; 
     } 
      return true; 
    } 

案例1：在只有1艺人的情况，则返回true监守没有其他艺术家比较

案例2：在2艺术家的情况下，与两个艺术家的物品返回true否则为false;

案例3：如果是3位艺术家，其中艺术家2和艺术家2的2个项目和艺术家3的5个项目，则返回false;

任何人都可以请这个我？

Answer 1

由于所有的艺术家需要有相同数量的材料......

function checkMaterials (arists) 
{ 
    if (!artists || !artists.length) { return false; } 
    if (artists.length < 2)   { return true; } 

    var valid = true; 
    var materialCount 

    try 
    { 
    //All artists must have the same number of materials, so we 
    //can test against the number of materials that the first 
    //artist has and reduce the number times we access the object 
    materialCount = (artists[0].materials.items || []).length; 
    } 
    catch (exception) 
    { 
    //Object is malformed 
    return false; 
    } 

    //Loop through the remaining artists and check how 
    //many materials they have against the first artist 
    for (var i = 1; i < artists.length; i++) 
    { 
    if (!artists[i].materials || ((artists[i].materials.items || []).length !== materialCount) 
    { 
     //Once one failed case is found, we can stop checking 
     valid = false; 
     break; 
    } 
    } 

    return valid; 
} 

//Test data 
var validArtists = [{ 
    materials: { 
    items: [1, 2, 3] 
    } 
}, { 
    materials: { 
    items: [1, 3, 4] 
    } 
}]; 

var invalidArtists = [{ 
    materials: { 
    items: [1, 2] 
    } 
}, { 
    materials: { 
    items: [3] 
    } 
}]; 

//Tests 
console.log (checkMaterials (validArsists)); //Prints true 
console.log (checkMaterials (invalidArtists)); //Prints false 

Answer 2

var artists = [{ 
 
\t materials: { 
 
\t \t items: [1, 2, 3] 
 
\t } 
 
}, { 
 
\t materials: { 
 
\t \t items: [1, 3] 
 
\t } 
 
}, { 
 
\t materials: { 
 
\t \t items: [1, 2, 3] 
 
\t } 
 
}, { 
 
\t materials: {} 
 
}]; 
 

 
artists.some(function(artist, i) { 
 
\t if (i === 0) return false; 
 
\t if (artists.length === 1) { 
 
\t \t console.log("Index " + i); 
 
\t \t console.log(true); 
 
\t \t return true; // length is one 
 
\t } 
 
\t if (artists[0].materials.items) { 
 
\t \t if (!artist.materials.items) { 
 
\t \t \t console.log("Index " + i); 
 
\t \t \t console.log(false); 
 
\t \t \t return false; // items doesn't exist. Return true/false, whatever works for you 
 
\t \t } else if (artist.materials.items && 
 
\t \t \t artist.materials.items.length === artists[0].materials.items.length) { 
 
\t \t \t console.log("Index " + i); 
 
\t \t \t console.log(true); 
 
\t \t \t return true; // length is equal 
 
\t \t } else { 
 
\t \t \t console.log("Index " + i); 
 
\t \t \t console.log(false); 
 
\t \t \t return false; // length is unequal 
 
\t \t } 
 
\t } else { 
 
\t \t if (artist.materials.items) { 
 
\t \t \t console.log("Index " + i); 
 
\t \t \t console.log(false); 
 
\t \t \t return false; // one has items, other doesn't 
 
\t \t } else { 
 
\t \t \t console.log("Index " + i); 
 
\t \t \t console.log(true); 
 
\t \t \t return true; // both have no items 
 
\t \t } 
 

 
\t } 
 
});

你为什么不尝试

artists.some(function(artist, i) { 
    if (i === 0) return false; 
    if (artists.length === 1) { 
     console.log("Index " + i); 
     console.log(true); 
     return true; // length is one 
    } 
    if (artists[0].materials.items) { 
     if (!artist.materials.items) { 
      console.log("Index " + i); 
      console.log(false); 
      return false; // items doesn't exist. Return true/false, whatever works for you 
     } else if (artist.materials.items && 
      artist.materials.items.length === artists[0].materials.items.length) { 
      console.log("Index " + i); 
      console.log(true); 
      return true; // length is equal 
     } else { 
      console.log("Index " + i); 
      console.log(false); 
      return false; // length is unequal 
     } 
    } else { 
     if (artist.materials.items) { 
      console.log("Index " + i); 
      console.log(false); 
      return false; // one has items, other doesn't 
     } else { 
      console.log("Index " + i); 
      console.log(true); 
      return true; // both have no items 
     } 

    } 
}); 

Answer 3

据我了解你只是想检查每个艺术家是否有相同数量的项目。此代码：

var result, materialsNumber; 
for (var artist of $scope.artists) { 
    var artistMaterialsNumber = artist.materials.items.length; 
    if (!materialsNumber) { 
     materialsNumber = artistMaterialsNumber; 
    } 
    result = (materialsNumber === artistMaterialsNumber); 
    if (!result) { 
     break; 
    } 
} 

return result; 

应该对此有用。它记得第一个艺术家的项目数量，并检查每个其他艺术家是否具有相同数量的项目。如果艺术家使用不同的商品编号代码中断并返回false。

Answer 4

要解决这个问题：

function checkValidity() { 
    var itemsCounts = $scope.artists.map(function(artist) { return artist.materials.items.length; }); 
    return itemsCounts.length > 1 
     ? itemsCounts.every(function(count) { return count === itemsCounts[0]; }) 
     : true; 
} 

Answer 5

可能是你可以做如下：

var artists = [{ materials: { items: [1, 2, 3] } }, 
 
       { materials: { items: [1, 2] } }, 
 
       { materials: { items: [] } }, 
 
       { materials: { items: [1] } }, 
 
       { materials: { items: [1, 2, 3] } } 
 
       ]; 
 
    result = artists.map(artist => artist.materials.items.length) 
 
        .every(length => length === artists[0].materials.items.length); 
 
console.log(result);

var artists = [{ materials: { items: [1, 2, 3] } } 
 
       ]; 
 
    result = artists.map(artist => artist.materials.items.length) 
 
        .every(length => length === artists[0].materials.items.length); 
 
console.log(result);

Answer 6

嗨，你也可以试试这个...

var vFirstItemLength = artists[0].materials.items.length; 
result = (artists.filter(function(item){return item.materials.items.length===vFirstItemLength;}).length === (artists.length)); 

Answer 7

解决方案：

function checkItemsValidity() { 
    if ($scope.artists.length === 1) { 
     return true; 
    } 

    for (var i = 0; i < $scope.artists.length; i++) { 
    //this condition might be unnecessary, I assumed items can be undefined from your code. 
    if (typeof $scope.artists[i].materials.items === 'undefined') { 
     $scope.artists[i].materials.items = []; 
    } 
    if (i === 0) { 
     continue; 
    } 
    if ($scope.artists[i].materials.items.length !== $scope.artists[0].materials.items.length) { 
     return false; 
    } 
    } 

    return true; 
} 

并与一些测试小提琴：https://jsfiddle.net/6x7zpkxe/1/