如果req.query.key为空，如何在db.find()方法中忽略它并返回所有文档

Question

我有两个来自pug模板的req.query参数：req.query.city和req.query.key

form.paieskaPagalRaktiniZodi(action='/search?' method='GET')
label(for='paieskaPagalRaktiniZodi') Key Word:
br
input#key(type='text' name='key')
br
label(for='city') City:
br
input#city(type='text' name='city')
br
button(type='submit' value='Submit' )

我想在我的DB上运行find方法，并查找包含给定值的文档，但问题是，如果我不在pug输入中插入任何内容，我会得到空字符串。如果我用空字符串运行find方法，我会得到0个结果。如果给定的查询为空，是否可以忽略find设置？

现在我用if's检查传入查询：

exports.filtruotiSkelbimai = catchAsync(async (req, res, next) => {
const keyName = req.query.key;
const cityName = req.query.city;
console.log(keyName);
let skelbimai;
if (keyName.length === 0) {
const skelbimaiSelected = await Skelbimai.find({
miestas: req.query.city,
});
skelbimai = skelbimaiSelected;
}
if (cityName.length === 0) {
const skelbimaiSelected = await Skelbimai.find({
$text: { $search: req.query.key },
});
skelbimai = skelbimaiSelected;
}
if (cityName.length > 0 && keyName.length > 0) {
const skelbimaiSelected = await Skelbimai.find({
$text: { $search: req.query.key },
miestas: req.query.city,
});
skelbimai = skelbimaiSelected;
}
if (keyName.length === 0 && cityName.length === 0) {
skelbimai = await Skelbimai.find();
}
res.status(201).render("overview", {
title: "Skelbimai",
skelbimai,
});
});

有可能做这样的吗

exports.filtruotiSkelbimai = catchAsync(async (req, res, next) => {
const keyName = req.query.key;
const cityName = req.query.city;
const skelbimai = await Skelbimai.find({
miestas: req.query.city,
$text: { $search: req.query.key },
});
res.status(201).render("overview", {
title: "Skelbimai",
skelbimai,
});
});

但是，如果任何给定的查询为空，则忽略该查询？如果两者都为空，则忽略它们，只需运行db.find()并返回所有文档。

尝试使用$或运算符执行此操作，但如果同时指示req.query.key和req.query.city，则它仅按关键字返回文档，而忽略城市。

Answer 1

您可以使用排列运算符以更干净的方式应用过滤器：

exports.filtruotiSkelbimai = catchAsync(async (req, res, next) => {
const keyName = req.query.key;
const cityName = req.query.city;
let filters = {};
if (cityName.length > 0) {
filters = {
...filters,
miestas: cityName,
};
}
if (keyName.length > 0) {
filters = {
...filters,
$text: { $search: keyName },
};
}
const skelbimai = await Skelbimai.find(filters);
res.status(201).render("overview", {
title: "Skelbimai",
skelbimai,
});
});