如果我想找到与"a"最接近的点,请使用序列化和地理空间查询.特定位置,Sequelize查询应如何?
Using Sequelize and geospatial queries, if I want to find the "n" closest points to a certain location, how should the Sequelize query be?
假设我有一个看起来像这样的模型:
Assume I have a model that looks something like this:
sequelize.define('Point', {geo: DataTypes.GEOMETRY('POINT')});现在假设我们通过类似以下方式在数据库中输入100个随机点:
Now let's say we input 100 random points in the db through something like:
db.Point.create({geo: {type: 'Point', coordinates: [randomLng,randomLat]}});想象一下,我们有一个lat和lng变量来定义一个位置,我们想找到最接近它的10个点.当我运行此查询时,我得到一个错误:
Imagine we have a lat and lng variables to define a location, and we want to find the 10 closest points to it. when I run this query I get an error:
const location = sequelize.literal(`ST_GeomFromText('POINT(${lat} ${lng})', 4326)`); db.Point.findAll({ attributes: [['distance', sequelize.fn('ST_Distance', sequelize.col('Point'), location)]], order: 'distance', limit: 10 }); // -> TypeError: s.replace is not a function任何想法是什么问题/如何解决?
Any idea what is the issue / how to fix it?
谢谢!
推荐答案用括号将 sequelize.fn 包围时,还必须包括一个字符串作为别名:
When you surround sequelize.fn with brackets, you must also include a string as an alias:
[sequelize.fn('ST_Distance_Sphere', sequelize.literal('geolocation'), location), 'ALIASNAME']此外,尝试将ST_Distance更改为ST_Distance_Sphere.所以:
Also, try changing ST_Distance to ST_Distance_Sphere. So:
const location = sequelize.literal(`ST_GeomFromText('POINT(${lng} ${lat})', 4326)`); User.findAll({ attributes: [[sequelize.fn('ST_Distance_Sphere', sequelize.literal('geolocation'), location),'distance']], order: 'distance', limit: 10, logging: console.log }) .then(function(instance){ console.log(instance); })这实际上对我有用. obs:确保用具有几何数据类型的模型替换用户".
This is actually working for me. obs: be sure you substitute 'User' with the model in which you have the geometry data type.
更新:如果您仍然无法使用order: 'distance'进行订购,也许您应该在var中声明它,并使用order: distance不带引号,例如:
Update: If you still can't order using order: 'distance', maybe you should declare it in a var and use order: distance without quotes, like this:
var lat = parseFloat(json.lat); var lng = parseFloat(json.lng); var attributes = Object.keys(User.attributes); var location = sequelize.literal(`ST_GeomFromText('POINT(${lng} ${lat})')`); var distance = sequelize.fn('ST_Distance_Sphere', sequelize.literal('geolocation'), location); attributes.push([distance,'distance']); var query = { attributes: attributes, order: distance, include: {model: Address, as: 'address'}, where: sequelize.where(distance, {$lte: maxDistance}), logging: console.log }距离精度更新:
sarikaya 提到的解决方案似乎更准确.这是使用postgres的方法:
The solution mentioned by sarikaya does seem to be more accurate. Here is how to do it using postgres:
var distance = sequelize.literal("6371 * acos(cos(radians("+lat+")) * cos(radians(ST_X(location))) * cos(radians("+lng+") - radians(ST_Y(location))) + sin(radians("+lat+")) * sin(radians(ST_X(location))))");更多推荐
序列化地理空间查询:找到"n".最接近某个位置的点
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