无法验证所有嵌套的子元素长度

Question

I have 1 object which contains nested child like below:

$scope.artists.materials.items[] 

Now i would have several artist which will contains list of items but in this i want to check total length of each item of artists and if mismatch found then i want to return true or false.

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

Idea here is to store the length of the items from the first artist and make sure all of them have that same items length.

Code:

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;
    }

Case 1 :In case of only 1 artist then return true becuase no other artist to compare

Case 2 : In case of 2 artist with 2 items for both artist return true else false;

Case 3 : In case of 3 artist with 2 items for artist1 and artist 2 and 5 items for artist3 then return false;

Can anybody please hlep me with this??

Answer 1

As I understand you just want to check if every artist has the same number of items. This code:

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;

should be useful for that. It remembers number of items of first artist and checks if every other artist has same number of items. In case there is artist with different item number code breaks and returns false .

Answer 2

Hi you can try this also...

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

Answer 3

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

Why don't you try

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 4

Since all artists need to have the same number of materials...

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 5

Should solve the issue:

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 6

May be you can do as follows;

 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 7

Solution:

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;
}

And a fiddle with some tests: https://jsfiddle.net/6x7zpkxe/1/