typeOrm 在 where 子句中使用条件

Question

@Injectable()

导出类 WeightPriceService { 构造函数(readonly dbContext: DbContext) {}

async findPriceByWeight(weight: number, tariffType?: PackageMaterialType): Promise<number> {
    const { price } = await this.dbContext.tariffs.findOne({
        where: {
            type: tariffType ? tariffType : ,
            isActive: true,
            weight: { min: LessThan(weight), max: MoreThan(weight) },
        },
        relations: ['weight'],
    });


    return price;
}

}

“tariffType”参数是否为 true 我想检查“type:tariffType”，否则不想检查“type”

Answer 1

您可以使用展开运算符来实现此目的。

const obj = {
  ...(true && {my: 'obj'})
};

const truly = 1 === 1;
const falsy = 1 !== 1;

const myObject = {
  foo: 'bar',
  ...(truly && {my: 'data'}),
  ...(falsy && {other: 'data'}),
  something: 'else'
};

console.log(myObject);

当条件为 True 时，它​​将注入对象，否则不会添加任何内容。

在你的情况下它会是

async findPriceByWeight(weight: number, tariffType?: PackageMaterialType): Promise<number> {
    const { price } = await this.dbContext.tariffs.findOne({
        where: {
            ...(tariffType && { type: tariffType }),
            isActive: true,
            weight: { min: LessThan(weight), max: MoreThan(weight) },
        },
        relations: ['weight'],
    });


    return price;
}