我有一个类似JSON的结构：
[
{
"name":"angelinas"
},
{
"name":"besuto"
},
{
"name":"catch",
"cuisine":"Japanese"
},
{
"name":"center cut"
},
{
"name":"fedora"
},
{
"name":"Habanero",
"cuisine":"Mexican"
},
{
"name":"Indies"
},
{
"name":"new"
},
{
"name":"RazINN"
},
{
"name":"restaurantTestVenue779"
},
{
"name":"restaurantTestVenue9703"
},
{
"name":"Salsa ",
"cuisine":"Mexican"
},
{
"name":"Sushi Place",
"cuisine":"Japanese"
},
{
"name":"The Ashoka"
},
{
"name":"The Poboys"
},
{
"name":"the shogun"
},
{
"name":"vinyard view"
}
]
使用上面的JSON，我想确定一种美食是否与餐厅相关联。如果是的话，我想构建一个类似JSON结构：
[
{
"Mexican":{
"venueNames":[
"Habanero",
"Salsa"
]
}
},
{
"Japanese":{
"venueNames":[
"Sushi Place",
"catch"
]
}
}
]
曾尝试使用for循环和.hasProperty构建JSON，但未成功。
答案 0 :(得分：1)
这是您可以做的！ 首先遍历数据，然后使用方法“ hasOwnProperty”检查该美食是否存在，是否存在，然后检查您的美食对象是否拥有该美食，然后将其添加到其中。
const data = [{
"name": "angelinas"
},
{
"name": "besuto"
},
{
"name": "catch",
"cuisine": "Japanese"
},
{
"name": "center cut"
},
{
"name": "fedora"
},
{
"name": "Habanero",
"cuisine": "Mexican"
},
{
"name": "Indies"
},
{
"name": "new"
},
{
"name": "RazINN"
},
{
"name": "restaurantTestVenue779"
},
{
"name": "restaurantTestVenue9703"
},
{
"name": "Salsa ",
"cuisine": "Mexican"
},
{
"name": "Sushi Place",
"cuisine": "Japanese"
},
{
"name": "The Ashoka"
},
{
"name": "The Poboys"
},
{
"name": "the shogun"
},
{
"name": "vinyard view"
}
]
let cuisines = {};
for (const resturant of data) {
if (resturant.hasOwnProperty('cuisine')) {
if (cuisines.hasOwnProperty(resturant.cuisine)) {
cuisines[resturant.cuisine].venueNames.push(resturant.name);
} else {
cuisines[resturant.cuisine] = {
venueNames: [resturant.name]
};
}
}
}
答案 1 :(得分：0)
这是对数组的简单简化。如果餐厅有已定义的美食，请检查结果是否已定义了该美食。如果没有，请为其创建一个对象，在其中您可以将餐厅名称推送至此。
const restaurants = [
{
"name":"angelinas"
},
{
"name":"besuto"
},
{
"name":"catch",
"cuisine":"Japanese"
},
{
"name":"center cut"
},
{
"name":"fedora"
},
{
"name":"Habanero",
"cuisine":"Mexican"
},
{
"name":"Indies"
},
{
"name":"new"
},
{
"name":"RazINN"
},
{
"name":"restaurantTestVenue779"
},
{
"name":"restaurantTestVenue9703"
},
{
"name":"Salsa ",
"cuisine":"Mexican"
},
{
"name":"Sushi Place",
"cuisine":"Japanese"
},
{
"name":"The Ashoka"
},
{
"name":"The Poboys"
},
{
"name":"the shogun"
},
{
"name":"vinyard view"
}
];
const cuisines = restaurants.reduce((result, restaurant ) => {
if ( restaurant.hasOwnProperty( 'cuisine' )) {
const { cuisine } = restaurant;
if ( !result.hasOwnProperty( cuisine )) {
result[ cuisine ] = {
venueNames: []
};
}
result[ cuisine ].venueNames.push( restaurant.name );
}
return result;
}, {});
console.log( cuisines );
我个人认为，我会使用略有不同的结构。如果我们用始终相同的对象表示集合，则可以简化大多数转换。这样效率不如一次循环完成所有操作，但是用于创建转换的代码几乎是可读的英语：
const restaurants = [
{ "name": "angelinas", "cuisine": null },
{ "name": "besuto", "cuisine": null },
{ "name": "catch", "cuisine": "japanese" },
{ "name": "center cut", "cuisine": null },
{ "name": "fedora", "cuisine": null },
{ "name": "habanero", "cuisine": "mexican" },
{ "name": "Indies", "cuisine": null },
{ "name": "new", "cuisine": null },
{ "name": "RazINN", "cuisine": null },
{ "name": "restaurantTestVenue779", "cuisine": null },
{ "name": "restaurantTestVenue9703", "cuisine": null },
{ "name": "Salsa ", "cuisine": "mexican" },
{ "name": "Sushi Place", "cuisine": "japanese" },
{ "name": "The Ashoka", "cuisine": null },
{ "name": "The Poboys", "cuisine": null },
{ "name": "the shogun", "cuisine": null },
{ "name": "vinyard view", "cuisine": null }
];
const create_cuisine = name => ({ name, "venues": [] });
const unique = () => {
const seen = {};
return item => {
const json = JSON.stringify( item );
return seen.hasOwnProperty( json )
? false
: ( seen[ json ] = true );
};
};
// Filter away all the restaurants without a cuisine value.
const restaurants_with_cuisine = restaurants.filter( restaurant => restaurant.cuisine );
const cuisines = restaurants_with_cuisine
// Extract the cuisine anmes from the restaurants.
.map( restaurant => restaurant.cuisine )
// Filter aways all the duplicates.
.filter( unique() )
// Create a new cuisine object.
.map( cuisine_name => create_cuisine( cuisine_name ));
// Finally add all the restaurant names to the right cuisine.
restaurants_with_cuisine.forEach( restaurant => cuisines.find( cuisine => cuisine.name === restaurant.cuisine ).venues.push( restaurant.name ));
console.log( cuisines );
答案 2 :(得分：0)
您可以在下面的一个循环中使用。
data.forEach(function(item) {
// if item has cuisine and cuisine not exist in new array
if(item["cuisine"] != null && typeof newArr.find(v => v[item.cuisine] != null) == 'undefined') {
// create new object with structure
let obj = {};
obj[item.cuisine] = {
"venueNames":[item.name]
};
newArr.push(obj);
}
else {
// else find existing cuisine and add new venue
let obj = newArr.find(v => v.hasOwnProperty(item.cuisine));
if(typeof obj != 'undefined') {
obj[item.cuisine].venueNames.push(item.name);
}
}
});
答案 3 :(得分：0)
使用一些es6功能，我们可以使用Set，map和filter生成此列表。
我们将首先映射一列美食，然后删除无效的美食，例如undefined。这样，我们将使用Set创建一个独特的美食列表。
接下来，我们将通过过滤与烹饪匹配当前迭代的原始对象，获取该列表并再次映射它以返回最终对象。最后，我们将过滤后的结果映射为仅将名称返回到venueNames对象。
我们的结果将如下所示：
function getItems(places) {
// Get a unique list of cuisines
return [...new Set(places.map(p => p.cuisine).filter(c => c))]
// Build the result
.map(c => {
return {
[c]: {
// Get a list of cuisines that match the current cuisine
venueNames: places.filter(p => p.cuisine == c).map(c => c.name)
}
}
})
}
const places = [
{"name": "angelinas"},
{"name": "besuto"},
{"name": "catch","cuisine": "Japanese"},
{"name": "center cut"},
{"name": "fedora"},
{"name": "Habanero","cuisine": "Mexican"},
{"name": "Indies"},
{"name": "new"},
{"name": "RazINN"},
{"name": "restaurantTestVenue779"},
{"name": "restaurantTestVenue9703"},
{"name": "Salsa ","cuisine": "Mexican"},
{"name": "Sushi Place","cuisine": "Japanese"},
{"name": "The Ashoka"},
{"name": "The Poboys"},
{"name": "the shogun"},
{"name": "vinyard view"}
]
console.log(getItems(places))