我有一个包含嵌套子项的对象，如下所示：
$scope.artists.materials.items[]
现在我会有几个艺术家将包含项目列表，但在此我想检查每个艺术家项目的总长度，如果发现不匹配，那么我想返回真或假。
问题是，当时我还没有任何艺术家的物品 我变得虚假
这里的想法是存储第一位艺术家的项目长度，并确保所有项目都具有相同的项目长度。
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 ：如果只有1位艺术家，则返回true，因为没有其他艺术家可以比较
案例2 ：如果2位艺术家同时为两位艺术家提供2项，则返回true，否则为false;
案例3 ：如果有3位艺术家，其中2位为artist1，艺术家为2，5位为artist3，则返回false;
有人可以用这个来发牢骚吗？
答案 0 :(得分：4)
据我了解，您只想检查每位艺术家是否拥有相同数量的商品。这段代码：
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。它记住第一位艺术家的项目数量，并检查每个其他艺术家是否有相同数量的项目。如果艺术家有不同的商品编号代码中断并返回false。
答案 1 :(得分：2)
您好，您也可以尝试一下......
var vFirstItemLength = artists[0].materials.items.length;
result = (artists.filter(function(item){return item.materials.items.length===vFirstItemLength;}).length === (artists.length));
答案 2 :(得分：1)
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
}
}
});
你为什么不试试
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
}
}
});
答案 3 :(得分：1)
由于所有艺术家都需要拥有相同数量的材料......
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
答案 4 :(得分：1)
应该解决问题：
function checkValidity() {
var itemsCounts = $scope.artists.map(function(artist) { return artist.materials.items.length; });
return itemsCounts.length > 1
? itemsCounts.every(function(count) { return count === itemsCounts[0]; })
: true;
}
答案 5 :(得分：1)
可能你可以这样做;
var artists = [{ materials: { items: [1, 2, 3] } },
{ materials: { items: [1, 2] } },
{ materials: { items: [] } },
{ materials: { items: [1] } },
{ materials: { items: [1, 2, 3] } }
];
result = artists.map(artist => artist.materials.items.length)
.every(length => length === artists[0].materials.items.length);
console.log(result);
var artists = [{ materials: { items: [1, 2, 3] } }
];
result = artists.map(artist => artist.materials.items.length)
.every(length => length === artists[0].materials.items.length);
console.log(result);
答案 6 :(得分：1)
解决方案：
function checkItemsValidity() {
if ($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;
}