Sequelize - where 子句中的列上的函数 [英] Sequelize - function on column in a where clause

问题描述

我有一个餐厅模型，其属性位置"的类型为 GEOMETRY("Point").我正在尝试使用 sequelize 编写一个查询，该查询将为我提供某个半径内的所有餐厅:

I have a Restaurant model with a property 'location' of the type GEOMETRY("Point"). I'm trying to write a query with sequelize that will give me all the restaurants in a certain radius:

models.Restaurant.findAll({
    attributes: ["*", [models.sequelize.fn("ST_Distance_Sphere", models.sequelize.fn("ST_MakePoint", latitude, longitude), models.sequelize.col("location")), "distance"]],
    where: {
        xxx: {
            $lte: radius
        }
    }
}).then(function(dishes) {
    res.status(200).json({dishes: dishes})
});

我不知道我应该在 xxx 的位置写什么.我尝试了各种方法，但每次都会出错.在选择查询中，我首先调用一个函数来从用户当前所在的坐标(纬度、经度)创建一个点.然后我使用一个函数来计算到餐厅位置的距离.
顺便说一句，我正在使用 postgres.

I have no clue what I should write where xxx is. I have tried all kinds of things but get an error every time. In the select query I'm first calling a function to create a point from the coordinates (latitude, longitude) where to user currently is. Then I use a function to calculate the distance to the location of the restaurant.
I'm using postgres btw.

最终这样做:

models.Restaurant.findAll({
    attributes: [
      'id',
      'name',
      'addressLine1',
      'addressLine2',
      'city',
      'zipPostalCode',
      'location',
      'phoneNumber',
      'website',
      [
        models.sequelize.fn(
          'ST_Distance_Sphere', models.sequelize.fn('ST_MakePoint', lat, long), models.sequelize.col('location')
        ),
        'distance',
      ],
    ],
    where: models.sequelize.and(
      models.sequelize.where(
        models.sequelize.fn(
          'ST_Distance_Sphere', models.sequelize.fn('ST_MakePoint', lat, long), models.sequelize.col('location')
        ), '>', fromRadius),
      models.sequelize.where(
        models.sequelize.fn(
          'ST_Distance_Sphere', models.sequelize.fn('ST_MakePoint', lat, long), models.sequelize.col('location')
        ), '<=', toRadius),
      models.sequelize.where(
        models.sequelize.col('Restaurant.is_online'), true
      )
    ),
  })

推荐答案

我假设 radius 是用户所在的计算点，xxx 应该是 distance

I'm assuming radius is that calculated point where the user is, xxx should be distance

检查doc，注意最后一个查询:>

Check the doc, notice the last query:

Model.findAll({
     where: sequelize.where(sequelize.fn('FUNCTION',
            sequelize.col('field')), 'value')
});

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