js数据处理问题，求解答或提供思路？

Question

筛选条件是按照类型分开渲染的，
这样写的

多条件筛选，这个怎么处理？

这是我的筛选条件

let conditionMapping = {
                coalName: this.coalList.filter(v => v.active == true).map(item => item.condition),
                orderType: this.typeList.filter(v => v.active == true).map(item => item.condition),
                customerName: this.customerList.filter(v => v.active == true).map(item => item.condition),
                xsUser2Name: this.forwarderList.filter(v => v.active == true).map(item => item.condition),
                fleetName: this.fleetList.filter(v => v.active == true).map(item => item.condition),
                filedName: this.filedList.filter(v => v.active == true).map(item => item.condition)
            };

            const mapping = {
                coalName: validateFn('coalName'),
                orderType: validateFn('orderType'),
                customerName: validateFn('customerName'),
                xsUser2Name: validateFn('xsUser2Name'),
                fleetName: validateFn('fleetName'),
                filedName: validateFn('filedName')
            };
            let list = this.taskList.map(val => val.taskCoalList.filter(v => Reflect.ownKeys(conditionMapping).every(key => mapping[key](v, conditionMapping[key]))));

            console.log(conditionMapping, '过滤条件');
            console.log(list, '过滤之后的数据');

Answer 1

先获取那些条件被选择了，因为被选择的条件有点多可以设置一个条件映射，以下是伪代码

const validateFn = (key) => (v, arr） => arr.includes(v[key])

// 建立如下条件映射
const mapping = {
    煤种: validateFn(煤种),
    车队: validateFn(车队),
    客户: validateFn(客户),
      ...
}

 
// 获取被选择的条件项并对conditionMapping赋值
// 例如被选择的条件映射如下
// 不管单选还是多选， value全都转化为数组为数组进行处理
let conditionMapping = {
    煤种：[3, 2, 6]，
    客户：[1008, 655],
    车队：[102]
}

const list = res.data.filter(v=> Reflect.ownKey(conditionMapping).every(key => mapping[key](v, conditionMapping[key]))) 

Answer 2

每个条件都可以多选,以类型为例:
假设你筛选的类型参数字段为type,多选的话type应该是个数组类型吧,假如当前筛选的是type=[1,2],那过滤出符合类型的条件的数据就是

res.data.filter(v=>v.type.includes(v.condictionType))

多个条件就后面继续 && 拼接