我正在尝试将id添加到数组中,如果它们可用,如下所示
export const extractOpaqueConstructionType = library => {
const opaqueConstructionSecondaryIds= [];
opaqueConstructionSecondaryIds.push(library?.exteriorWallId);
opaqueConstructionSecondaryIds.push(library?.exteriorFloorId);
opaqueConstructionSecondaryIds.push(library?.roofId);
opaqueConstructionSecondaryIds.push(library?.interiorWallId);
opaqueConstructionSecondaryIds.push(library?.interiorFloorId);
opaqueConstructionSecondaryIds.push(library?.belowGradeWallId);
opaqueConstructionSecondaryIds.push(library?.slabOnGradeId);
return { opaqueConstructions: opaqueConstructionSecondaryIds || null };
};我在其他地方调用上面的函数,如下所示
extractSecondaryIds: library => {
const secondaryIds = {
...extractSourceOfData(library),
...extractOpaqueConstructionType(library), // this is where i am calling above function
...extractGlazingConstructionType(library)
};
return secondaryIds;
},例如,如果上面的所有id都是(exteriorWallId、exteriorFloorId等)都是未定义的,我从上面的函数(extractOpaqueConstructionType)得到输出,比如opaqueConstructions: [null],如果id是未定义的,我希望是这样的opaqueConstructions: null
请任何人在这方面提出任何想法或替代方法,非常感谢我,非常感谢提前。
发布于 2020-12-09 03:43:19
推送完所有值后,需要对数组进行filter。
您还可以通过简单地删除推式调用来简化代码
export const extractOpaqueConstructionType = library => {
const opaqueConstructions = [
library?.exteriorWallId,
library?.exteriorFloorId,
library?.roofId,
library?.interiorWallId,
library?.interiorFloorId,
library?.belowGradeWallId,
library?.slabOnGradeId,
].filter(it => it !== null);
return { opaqueConstructions: opaqueConstructions.length ? opaqueConstructions : null };
}发布于 2020-12-09 03:03:03
尝试如下所示
export const extractOpaqueConstructionType = library => {
const opaqueConstructionSecondaryIds= [
library?.exteriorWallId
library?.exteriorFloorId,
library?.roofId,
library?.interiorWallId,
library?.interiorFloorId,
library?.belowGradeWallId,
library?.slabOnGradeId,
].filter(keyValue => keyValue!==undefine);
return { opaqueConstructions:opaqueConstructionSecondaryIds.length?
opaqueConstructionSecondaryIds : null };
};https://stackoverflow.com/questions/65204124
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