Question

我有两张桌子。一个是类别表，另一个是具有类别外键的项目表（一个类别可以有多个项目）

我试图公开以下代码中的所有项目

router.route('/projects')
  .get(function(req,res){
    //knex('projects').select(['projects.id','projects.name','projects.category_id','categories.name']).innerJoin('categories','projects.category_id','categories.id')
    knex.from('projects').innerJoin('categories','projects.category_id','categories.id')
    .then(function(collection){

      console.log(collection);
      res.json({
        error:false,
        data: collection
      })
    })
    .catch(function(err){
      res.json({
        error:true,
        data:{
          message:err.message
        }
      })
    })
  })

两个knex语句都返回了类似于下面的截断输出

{
  "error": false,
  "data": [
    {
      "id": 1,
      "name": "Music",
      "category_id": 1,
      "created_at": 1458126413858
    },
    {
      "id": 2,
      "name": "Science",
      "category_id": 2,
      "created_at": 1458126413858
    }
  ]
}

但是，原始SQL似乎是正确的

select * from "projects" inner join "categories" on "projects"."category_id" = "categories"."id"

或

select "projects"."id", "projects"."name", "projects"."category_id", "categories"."name" from "projects" inner join "categories" on "projects"."category_id" = "categories"."id"

在sqlite上单独运行上述语句似乎返回整个表结果

knex声明在哪里出错？

供参考，这是架构

项目表

var table = (table) => {
  table.increments().primary();
  table.string('name');
  table.integer('category_id').references('categories.id');
  table.timestamp('created_at').defaultTo(Date.now());
}

类别表

var table = (table) => {
  table.increments().primary();
  table.string('name');
  table.timestamp('created_at').defaultTo(Date.now());
}

Answer 1

通过使用columnNames的别名来解决它，如下所示

knex('projects').select(['projects.id as projectId','projects.name as projectName','projects.category_id','categories.name as categoryName']).innerJoin('categories','projects.category_id','categories.id')

Knex Js从原始SQL返回不同的输出

1 个答案: