在数组java脚本中使用2过滤器

Question

我有一个类似的数组

我只想带支票=真和支票=真的分期付款

我有这个阵列

0: {
checked: true,
code: 100.
paymant: Array(2)
{
0: {data: 03/12/2021, checked: true},
1: {data: 03/12/2021}
}
},
1: {
code: 100.
paymant: Array(2)
{
0: {data: 03/12/2021},
1: {data: 03/12/2021}
}
}
// Real Data
const data = [
{
checked: true,
code: 100,
payments: [
{data: 'checked', checked: true},
{data: 'not-checked'}
]
},
{
code: 100,
payments: [
{data: 'not-checked'},
{data: 'not-checked'}
]
}
];

我想要这个

0: {
checked: true,
code: 100.
paymant: Array(1)
{
0: {data: 03/12/2021, checked: true},
}
},

我试着这样做，但没有工作

installments.filter(i => i.checked).filter(i => i.paymant.filter(p => p.checked))

Answer 1

第一个过滤器是直接的，但您需要重新构建支付数组，使其只包含选中的条目。

const data = [
{
checked: true,
code: 100,
payments: [
{data: 'checked', checked: true},
{data: 'not-checked'}
]
},
{
code: 100,
payments: [
{data: 'not-checked'},
{data: 'not-checked'}
]
}
];
const checked = (item) => item && item.checked;
const paymentCheckedMapper = (item) => (
{...item, payments: item.payments.filter(checked)}
);
data.filter(checked).map(paymentCheckedMapper)

https://jsfiddle.net/vebwfnL0/

Answer 2

因为您需要混合和匹配filter和map。您可以在此处使用reduce无需过滤然后映射，这将带来性价比问题。

const data = [
{
checked: true,
code: 100,
paymant: [
{ data: 03 / 12 / 2021, checked: true },
{ data: 03 / 12 / 2021 },
],
},
{
code: 101,
paymant: [{ data: 03 / 12 / 2021 }, { data: 03 / 12 / 2021 }],
},
];
const getPaid = (records = []) => {
return records.reduce((acc, record) => {
if (!record.checked) return acc;
const paymant = record.paymant.filter((paymant) => paymant.checked);
if (paymant.length) {
acc.push({ ...record, paymant });
}
return acc;
}, []);
};
console.log(getPaid(data));