使用JavaScript从JSON文件中分离多个数据
我试图使用以下代码显示从API获取的JSON文件中的值使用JavaScript从JSON文件中分离多个数据,javascript,json,Javascript,Json,我试图使用以下代码显示从API获取的JSON文件中的值 const query = 'chicken'; const uri = 'https://trackapi.nutritionix.com/v2/search/instant?query=' + query; let appid = new Headers(); appid.append('x-app-id', '19421259') let appkey = new Headers(); appid.append('x-app-key
const query = 'chicken';
const uri = 'https://trackapi.nutritionix.com/v2/search/instant?query=' + query;
let appid = new Headers();
appid.append('x-app-id', '19421259')
let appkey = new Headers();
appid.append('x-app-key', '54a4e6668518084478d9025d8a5e32a2')
let req = new Request(uri, {
method: 'GET',
headers: appid,
appkey,
});
fetch(req)
.then((response) => {
if (response.ok) {
return response.json();
} else {
throw new Error('BAD HTTP stuff');
}
})
.then((jsonData) => {
console.log(jsonData);
var jsonString = JSON.stringify(jsonData);
document.write(jsonString);
})
.catch((err) => {
console.log('ERROR', err.message);
});
它成功返回JSON并将其转换为字符串,但我不知道如何将它们分开,因为JSON包含多个数据(例如,搜索鸡肉会返回多个不同品牌和烹饪类型的鸡肉餐),如下所示:
{
"common": [{
"food_name": "chicken",
"serving_unit": "oz",
"tag_name": "chicken",
"serving_qty": 3,
"common_type": null,
"tag_id": "9",
"photo": {
"thumb": "https://d2xdmhkmkbyw75.cloudfront.net/9_thumb.jpg"
},
"locale": "en_US"
}, {
"food_name": "chickensalad",
"serving_unit": "cup",
"tag_name": "chicken salad",
"serving_qty": 0.5,
"common_type": null,
"tag_id": "1420",
"photo": {
"thumb": "https://d2xdmhkmkbyw75.cloudfront.net/1420_thumb.jpg"
},
"locale": "en_US"
}, {
"food_name": "chicken salad",
"serving_unit": "cup",
"tag_name": "chicken salad",
"serving_qty": 0.5,
"common_type": null,
"tag_id": "1420",
"photo": {
"thumb": "https://d2xdmhkmkbyw75.cloudfront.net/1420_thumb.jpg"
},
"locale": "en_US"
}, {
"food_name": "chicken broth",
"serving_unit": "cup",
"tag_name": "broth chicken",
"serving_qty": 1,
"common_type": null,
"tag_id": "3336",
"photo": {
"thumb": "https://d2xdmhkmkbyw75.cloudfront.net/3336_thumb.jpg"
},
"locale": "en_US"
}, {
"food_name": "whole chicken",
"serving_unit": "chicken",
"tag_name": "whole chicken",
"serving_qty": 1,
"common_type": null,
"tag_id": "4025",
"photo": {
"thumb": "https://d2xdmhkmkbyw75.cloudfront.net/4025_thumb.jpg"
},
"locale": "en_US"
}
我应该如何实现它,以便能够分别显示每种食物及其各自的变量?如果我理解正确,您将尝试过滤作为JSON对象获取的数据,以便它只包括“食物名称”和“标签名称”属性中保存的食物 在这种情况下,您可以编写一个简单的函数来进行过滤,如下所示:
const getFilteredFoodsArray = food => {
food = food.toLowerCase();
return common.filter(
data =>
data.food_name.toLowerCase().includes(food) ||
data.tag_name.toLowerCase().includes(food)
);
};
然后,如果您这样运行它:
console.log(getFilteredFoodsArray("chicken"))
你的结果是:
[
{ food_name: 'chicken',
serving_unit: 'oz',
tag_name: 'chicken',
serving_qty: 3,
common_type: null,
tag_id: '9',
photo:
{ thumb: 'https://d2xdmhkmkbyw75.cloudfront.net/9_thumb.jpg' },
locale: 'en_US' },
{ food_name: 'chickensalad',
serving_unit: 'cup',
tag_name: 'chicken salad',
serving_qty: 0.5,
common_type: null,
tag_id: '1420',
photo:
{ thumb: 'https://d2xdmhkmkbyw75.cloudfront.net/1420_thumb.jpg' },
locale: 'en_US' },
{ food_name: 'chicken salad',
serving_unit: 'cup',
tag_name: 'chicken salad',
serving_qty: 0.5,
common_type: null,
tag_id: '1420',
photo:
{ thumb: 'https://d2xdmhkmkbyw75.cloudfront.net/1420_thumb.jpg' },
locale: 'en_US' },
{ food_name: 'chicken broth',
serving_unit: 'cup',
tag_name: 'broth chicken',
serving_qty: 1,
common_type: null,
tag_id: '3336',
photo:
{ thumb: 'https://d2xdmhkmkbyw75.cloudfront.net/3336_thumb.jpg' },
locale: 'en_US' },
{ food_name: 'whole chicken',
serving_unit: 'chicken',
tag_name: 'whole chicken',
serving_qty: 1,
common_type: null,
tag_id: '4025',
photo:
{ thumb: 'https://d2xdmhkmkbyw75.cloudfront.net/4025_thumb.jpg' },
locale: 'en_US' }
]
现在还不清楚你希望结果是什么样子。你能举一个例子,说明鸡肉项目经过转换后应该是什么样的吗?很抱歉,但我想要的是以列表形式显示鸡肉项目,显示其详细信息,因为这将在搜索功能中实现,用户将搜索特定食物,并显示与其相关的所有食物搜索我不确定我在这里不了解的内容-此数据看起来可以通过迭代
common
来显示。很抱歉,我可能没有正确解释,但我想做的是从结果中获取每个数据(例如,仅获取此>{食物名称:'chicken',上菜单位:'oz',标签名称:'chicken',上菜数量:3,常见类型:null,标签id:'9',照片:{拇指:'},区域设置:'en_US'},这样我就可以单独显示它们了。我是一个使用json的初学者,所以这可能是一个非常基本的东西,我无法理解