我有一个对象，它具有三个不同的数组，例如位置垂直和圆形类型，我将获得一个过滤器对象，该对象将在该对象中具有相同的三个数组。这是需要过滤的数据testObject = [{ "id": 1892928, "vertical_tax": [ 678, 664 ], "location_tax": [ 666 ], "roundType": [ "rt1" ],}{ "id": 1892927, "vertical_tax": [ 662, 663 ], "location_tax": [ 663 ], "roundType": [ "rt2" ],}]这是过滤应该基于的过滤器对象 filterObject = { locations: [666,667] roundTypes: ["rt1","rt2"] verticals: [662,661] }原始要求：是在任一过滤器对象数组中获取具有特定值的任何对象。这可以通过使用“一些”来完成。更新要求：所以我需要使用 filterObject 中传递的值来过滤主对象。所以filterObject中的所有条件都应该匹配。应该返回所有匹配的 id。这可以用“every”来完成
2 回答
至尊宝的传说
TA贡献1789条经验 获得超10个赞
您可以过滤对象并排除它们。我已经注释掉了部分比较，因为不清楚是否要过滤这些属性以及如何过滤。你只提到了地点。如果您希望它包含所有属性的所有匹配结果，&&请将||.
如前所述，如果属性匹配（或具有一致的命名），则可以简化和概括代码。
filterObjects过滤存在的任何匹配项。
filterObjects1要求存在 中的所有元素verticals，以及其他属性中的任何匹配项。
testObject = [{
"id": 1892928,
"vertical_tax": [
678,
664
],
"location_tax": [
666
],
"roundType": [
"rt1"
],
},
{
"id": 1892927,
"vertical_tax": [
662,
663
],
"location_tax": [
663
],
"roundType": [
"rt2"
],
}]
filterObject = {
locations: [666,667],
roundTypes: ["rt1","rt2"],
verticals: [662,661]
};
const filterObjects = (filterObject, testObject) => {
return testObject.filter(obj =>
obj.location_tax && obj.location_tax.some(
x => filterObject.locations && filterObject.locations.includes(x)) ||
obj.roundType && obj.roundType.some(
x => filterObject.roundTypes && filterObject.roundTypes.includes(x)) ||
obj.vertical_tax && obj.vertical_tax.some(
x => filterObject.verticals && filterObject.verticals.includes(x))
);
};
console.log(filterObjects(filterObject, testObject));
filterObject = {
roundTypes: ["rt1","rt2"],
verticals: [662,661]
};
console.log(filterObjects(filterObject, testObject));
// require presence of all objects in filterObject.verticals using .every
const filterObjects1 = (filterObject, testObject) => {
return testObject.filter(obj =>
(obj.location_tax && obj.location_tax.some(
x => filterObject.locations && filterObject.locations.includes(x)) ||
obj.roundType && obj.roundType.some(
x => filterObject.roundTypes && filterObject.roundTypes.includes(x))
) &&
filterObject.verticals.every( x => obj.vertical_tax && obj.vertical_tax.includes(x) )
);
};
filterObject = {
roundTypes: ["rt1","rt2"],
verticals: [662,663]
};
delete testObject[0].vertical_tax;
console.log(filterObjects1(filterObject, testObject));
//必须匹配所有存在的filterObject 属性中的某个值
testObject = [{
"id": 1892928,
"vertical_tax": [
678,
664
],
"location_tax": [
666
],
"roundType": [
"rt1"
],
},
{
"id": 1892927,
"vertical_tax": [
662,
663
],
"location_tax": [
663
],
"roundType": [
"rt2"
],
}]
filterObject = {
locations: [666,667],
roundTypes: ["rt1","rt2"],
// verticals: [662,661,678]
};
// _must_ match some value in _all_ filterObject properties _that exist_
const filterObjects = (filterObject, testObject) => {
return testObject.filter(obj =>
(!filterObject.locations || obj.location_tax && obj.location_tax.some(
x => filterObject.locations && filterObject.locations.includes(x))) &&
(!filterObject.roundTypes || obj.roundType && obj.roundType.some(
x => filterObject.roundTypes && filterObject.roundTypes.includes(x))) &&
(!filterObject.verticals || obj.vertical_tax && obj.vertical_tax.some(
x => filterObject.verticals && filterObject.verticals.includes(x)))
);
};
console.log(filterObjects(filterObject, testObject));
慕码人2483693
TA贡献1860条经验 获得超9个赞
如果您只想测试一个条件，只需使用该语句并丢弃其余条件，或者如果您想让任何一个条件为真以获得结果，则将 && 逻辑更改为 || 必要的陈述。
testObject.filter( i => {
return i.vertical_tax.every((value, index) => value === filterObject.verticals[index]) &&
i.location_tax.every((value, index) => value === filterObject.locations[index]) &&
i.roundType.every((value, index) => value === filterObject.roundTypes[index]);
});
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