Vit*_*lyk 4 javascript postgresql postgis node.js sequelize.js
我在搜索时遇到了问题,包括按某个点的距离排序。这是我的代码以及我要执行的操作。感谢帮助
const Sequelize = require('sequelize');
var Flat = db.define('flat', {
id: {
type: Sequelize.INTEGER,
autoIncrement: true,
primaryKey: true
}
});
var FlatAddress = db.define('flat_address', {
id: {
type: Sequelize.INTEGER,
autoIncrement: true,
primaryKey: true
},
flat_id: {
type: Sequelize.INTEGER,
foreignKey:true,
allowNull:false,
references: {
model:'flats',
key: 'id'
}
},
city: {
type: Sequelize.STRING(50) //post_town
},
location: {
type: Sequelize.GEOMETRY('POINT')
}
});
Flat.hasOne(FlatAddress, { as: 'Address', foreignKey: 'flat_id', otherKey: 'id', onDelete: 'cascade' });
FlatAddress.belongsTo(Flat, { foreignKey: 'id', otherKey: 'flat_id', onDelete: 'cascade' });
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我想做这样的事情
var POINT = {lat, lng} ??
Flats.findAndCountAll({
where: filter,
order: [
[ { model: FlatAddresses, as: 'Address' },
'//here should be something like distance from POINT//', 'ACS']
],
include: [
{ model: FlatAddresses, as: 'Address'}
],
offset,
limit
})
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我没有找到适合我的案例的示例或文档。谢谢
以下语句将Flats在您的经度和纬度内找到,distance在有效负载中包含一个称为的字段,并按进行排序distance。
const myDistance = 10000; // e.g. 10 kilometres
Flats.findAll({
attributes: {
include: [
[
Sequelize.fn(
'ST_Distance',
Sequelize.col('location'),
Sequelize.fn('ST_MakePoint', longitude, latitude)
),
'distance'
]
]
},
where: Sequelize.where(
Sequelize.fn(
'ST_DWithin',
Sequelize.col('location'),
Sequelize.fn('ST_MakePoint', longitude, latitude),
myDistance
),
true
),
order: Sequelize.literal('distance ASC')
});
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请记住:我不知道您使用的是哪个SRID。因此,尽管您可能需要在代码中将米转换为半径,但我使用米来测量距离。
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