node.js - 通过多种条件获取资源的最佳实践是什么？

Question

使用 Express.js 和 MongoDB，我想在没有带有请求 url 的查询参数时获取所有教师列表，我按照如下条件执行操作，例如找到查询时然后执行此操作，否则执行此操作，是否高效？或者有什么好的建议。提前致谢

export const getTeachers = (req, res, next) => {
    if(req.query.course_id) {
        try {
            Teacher.find({'course_ids':req.query.course_id}).exec()
                .then((teachers) => {
                    if (!teachers) {
                        return next(errorPlaceHolder(403, false, 'no teacher found'));
                    }
                    return res.status(200).json({
                        success: true,
                        message: 'Read course teacher',
                        teachers: teachers
                    })
                })
                .catch((error) => {
                    next(error);
                });
        } catch (e) {
            res.status(500).json({
                success: false,
                message: 'Error in get Teachers',
                error: e
            });
        }
    }else{
        try {
            Teacher.find().exec()
                .then((teachers) => {
                    if (!teachers) {
                        return next(errorPlaceHolder(403, false, 'no teacher found'));
                    }
                    return res.status(200).json({
                        success: true,
                        message: 'Read all teacher',
                        teachers: teachers
                    })
                })
                .catch((error) => {
                    next(error);
                });
        } catch (e) {
            res.status(500).json({
                success: false,
                message: 'Error in get Teachers',
                error: e
            });
        }
    }
}

Answer 1

你可以这样优化代码

var findCriteria = {}
if(req.query.course_id) {
   findCriteria['course_ids'] = req.query.course_id;
}

Teacher.find(findCriteria).exec()
                    .then((teachers) => {

 })