本文介绍了我怎样才能基于多个条件过滤我的数据?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
好的,我有一个对象数组,这是我要过滤的数据:
const posts = [
{
hairday: '2',
products: ['product1', 'product2', 'product3', 'product4'],
rating: '2',
drying: 'diffuser'
},
{
hairday: '2',
products: ['product6', 'product7', 'product8', 'product9'],
rating: '4',
drying: 'air dry'
},
{
hairday: '3',
products: ['product10', 'product11', 'product13', 'product14'],
rating: '3',
drying: 'air dry'
},
{
hairday: '4',
products: ['product15', 'product26', 'product37', 'product14'],
rating: '5',
drying: 'towel dry'
},
]
我想通过多个条件过滤上述数据。以下是条件的示例对象:
{
products: ['product1', 'product13'],
hairday: ['2', '3'],
drying: ['air dry', 'diffuser'],
rating: []
}
所以我要获取与每个数组中至少一项匹配的所有post
对象。
因此,过滤项目应具有Product1或Product13和Hairday 2或Hairday 3以及风干或烘干扩散器和任何额定值。
解决此问题的最佳方式是什么?我的过滤对象的结构是最好的吗? 提前感谢<;3
推荐答案
对于不需要手动列出键的更健壮的解决方案,您可以使用Object.keys()
迭代您的Criteria对象,然后将post
数组中的各个对象与其进行比较:
const filtered = posts.filter(post => {
const conditionKeys = Object.keys(conditions);
const fulfillments = conditionKeys.map(k => {
const condition = conditions[k];
// If we encounter an empty array, then criteria is always met
if (condition.length === 0) {
return true;
}
// Enforces that as long as ONE subcondition is met (`OR`)
return condition.filter(v => post[k].includes(v)).length > 0;
});
// Enforce that EVERY condition must be met (`AND`)
// A condition is considered met as long as it is true
return fulfillments.every(x => x);
});
在迭代各个条件(Hairday、Products等)时,我们只是想检查您的对象是否包括一个或多个列出的值(即两个数组中的任何一个都是相交的)。如果有交叉点,则长度为>;0,否则为0。
请参阅下面的概念验证:
const posts = [{
hairday: '2',
products: ['product1', 'product2', 'product3', 'product4'],
rating: '2',
drying: 'diffuser'
},
{
hairday: '2',
products: ['product6', 'product7', 'product8', 'product9'],
rating: '4',
drying: 'air dry'
},
{
hairday: '3',
products: ['product10', 'product11', 'product13', 'product14'],
rating: '3',
drying: 'air dry'
},
{
hairday: '4',
products: ['product15', 'product26', 'product37', 'product14'],
rating: '5',
drying: 'towel dry'
},
];
const conditions = {
products: ['product1', 'product13'],
hairday: ['2', '3'],
drying: ['air dry', 'diffuser'],
rating: []
};
const filtered = posts.filter(post => {
const fulfillments = Object.keys(conditions).map(k => {
const condition = conditions[k];
// If we encounter an empty array, then criteria is always met
if (condition.length === 0) {
return true;
}
return condition.filter(v => post[k].includes(v)).length > 0;
});
return fulfillments.every(x => x);
});
console.log(filtered);
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