MySql-字段“ park_id”没有默认值

Question

I am very new to Mysql, and I have been stuck on this for a while.

I am using bookshelfjs to create a many-to-many relatioship,everytime I try to attach my articles to parks I get the following error.

(node:24480) UnhandledPromiseRejectionWarning: Unhandled promise rejection (rejection id: 1): Error: insert into articles_parks ( article_id , park_id ) values (0, DEFAULT) - ER_NO_DEFAULT_FOR_FIELD: Field 'park_id' doesn't have a default value

this is my MySQL code:

Create table parks( 
   park_id integer,
   primary key (park_id)
);

Create table articles(
   article_id integer,
   primary key (article_id)
);

CREATE TABLE articles_parks(
   article_id integer,
   park_id    integer,
   Primary Key (article_id,park_id), 
   Foreign Key (article_id) REFERENCES articles(article_id),
   Foreign Key (park_id) REFERENCES parks(park_id)
);

this is my bookshelf code:

var bookshelf = require('bookshelf')(knex);

    var Article = bookshelf.Model.extend({
      tableName: 'articles',
      park: function() {
        return this.belongsToMany(Park);
      }
    });

    var Park = bookshelf.Model.extend({
      tableName: 'parks',
      articles: function() {
        return this.belongsToMany(Article);
      }
    });
    const number = 197;
    var article1 = new Article({article_id: number});

    Promise.all([article1.save()])
      .then(function() {
        return Promise.all([
        new Park({park_id: 200}).articles().attach([article1]),
      ]);
    })

Answer 1

Your two primary keys (park_id and article_id) are not set to be auto incremented. In MySQL, you must explicitly state that you want the primary keys to be auto incremented or it will assume you are going to manually assign them values. This is why it is returning an error that they are missing a default value.

Try editing your SQL code to this:

Create table parks( 
   park_id integer AUTO_INCREMENT,
   primary key (park_id)
);

Create table articles(
   article_id integer AUTO_INCREMENT,
   primary key (article_id)
);

CREATE TABLE articles_parks(
   article_id integer,
   park_id    integer,
   Primary Key (article_id,park_id), 
   Foreign Key (article_id) REFERENCES articles(article_id),
   Foreign Key (park_id) REFERENCES parks(park_id)
);