我正在尝试分解用户的获取请求并将其放入名为查询的变量中。然后使用它的where子句将var查询传递到sequelize的findAll方法中,看起来Sequelize认为我正在寻找一个表CALLED查询,而实际上我正在尝试传递对象。如果我不能很好地解释,我很抱歉,但这是代码和错误:
var info = [];
//link example: localhost:8081/filter/?descripiton=san+francisco&houseType=house&numOfBedroom=3&numOfBathroom=2&houseSize=500&price=1200
exports.filterListings = function(req) {
//create an object literal which we will return, and has a nested object named filteredList inside.
//filteredList contains an array named listings where we will put listings that match our filter inside
let response = {
filteredList: {listings: []},
};
//now we need to see how the user wants us to filter the listings
const query = req.query;
//do some logic where we decompose query
if(query.descripiton != undefined) {
//info = info + 'descripiton: ' + query.descripiton+', ';
info.push('descripiton: ' + query.descripiton+', ');
console.log(info);
}
if(query.houseType != undefined) {
//info = info + 'houseType: ' + query.houseType+', ';
info.push('houseType: ' + query.houseType+', ');
//console.log(info);
}
if(query.numOfBedroom != undefined) {
//info = info + 'numOfBedroom: ' + query.numOfBedroom+', ';
info.push('numOfBedroom: ' + query.numOfBedroom+', ');
}
if(query.numOfBathroom != undefined) {
//info = info + 'numOfBathroom: ' + query.numOfBathroom+', ';
info.push('numOfBathroom: ' + query.numOfBathroom+', ');
}
if(query.houseSize != undefined) {
//info = info + 'houseSize: ' + query.houseSize+', ';
info.push('houseSize: ' + query.houseSize+', ');
}
if(query.price != undefined) {
//info = info + 'price: ' + query.price;
info.push('price: ' + query.price);
}
然后当我尝试传递信息变量时
listingModel.findAll({
//error because it wont recognize the variable search nor will it recognize info
where: {info}
}).then(listings => {
// so we loop through listings and insert what we have found into the response (which we are going to return)
for(var i = 0; i < listings.length; i++) {
response.filteredList.listings.push(listings[i]);
}; // loop where we insert data into response done
我希望它根据动态查询查找所有列表,但我收到错误:
Unhandled rejection SequelizeDatabaseError: Unknown column 'Listing.info' in 'where clause'
非常感谢您提供的潜在帮助!
最佳答案
让我们尝试将您的问题一一进行排序。抱歉,双关语:p
而不是使用多个 if
来创建过滤列表。使用for ... in
。然后使用该对象数组以及 Sequelize.Op
来创建查询。
示例:
const Op = require('sequelize').Op;
const whereClause = [];
const query = req.query;
for(const key in query) {
if(query[key] !== '' && query[key] !== null) {
//object will be pushed to the array like "houseType:big"
whereClause.push({key:query[key]})
}
}
//you now have the where clause
//use it in your query with Op.and
listingModel.findAll({
where: {
[Op.and]: whereClause,
}
});
有关使用 Sequelize - Operators 查询的更多信息
关于javascript - 在 where 子句中传递动态查询时出现未处理的拒绝 SequelizeDatabaseError,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56011859/