Question

通过将function作为参数传递给Model.find()->这样的Model.find(searchCondition)，我在查询MongoDB（猫鼬）时遇到问题。我希望你能帮助我。

// Fetching patients from the database
exports.getPatients = (req, res, next) => {
    const criterionSearchCategory = req.query.kriterijumPretrage;
    const ageSearchCategory = req.query.kriterijumGodina;
    const searchInputValue = req.query.pojamPretrage;

    console.log({ [criterionSearchCategory]: { [ageSearchCategory]: Number([searchInputValue]) }});

    // Patient search condition, based on selected option from select dropdown
    function searchCondition() {
        if (criterionSearchCategory == 'undefined') {
            return {};
        } else if (criterionSearchCategory == 'age') {
            return { [criterionSearchCategory]: { [ageSearchCategory] : Number([searchInputValue]) }}
        } else {
            return { [criterionSearchCategory]: { $in: [ "/^" + searchInputValue + "/i" ]}}
        }
    }

   ...

    const patientQuery = Patient.find(searchCondition);

getPatients(patientsPerPage: number, currentPage: number, criterionSearchCategory: string, searchInputValue: string, ageSearchCategory: any) {
        const queryParams = `?pacijenataPoStranici=${patientsPerPage}&trenutnaStranica=${currentPage}&kriterijumPretrage=${criterionSearchCategory}&pojamPretrage=${searchInputValue}&kriterijumGodina=${ageSearchCategory}`;
        this.http
            .get<{ message: string, patients: any, maxPatients: number }>( BACKEND_URL + queryParams)
            // Execute map on every data that makes it through Observable stream
            .pipe(map((patientData) => {

我想在手动传递查询参数时进行修饰，例如const patientQuery = Patient.find({ age: { '$gt': 30 } });将正确地提取适当的患者，但是当我传递函数时，例如const patientQuery = Patient.find(searchCondition);然后不起作用。

第一个问题，是否可以像这样将函数作为参数传递？

任何建议将不胜感激。谢谢

MongoDB查询-将函数传递给Model.find（）

0 个答案: