relationship

laravel 5.4 many to many relationship

由 时间秒杀一切 提交于 2019-12-10 10:45:41
问题 i have relationship many to many table 'admins' , 'pjt_roles' with pjt_role_admin. but,not working i have 2 model class Role protected $table = 'pjt_roles'; public function Admin(){ return $this->belongsToMany(Admin::class',pjt_role_admin'); } class Admin public function Role(){ return $this->belongsToMany(Role::class,'pjt_role_admin'); } and table pjt_role_admin have attribute admin_id from table admins role_id from table pjt_roles 回答1: Specify your pivot table in relationship. Default

how to filter records filtered by more than one criterion in pivot table - laravel eloquent collections

由 余生颓废 提交于 2019-12-10 10:42:41
问题 I am remodeling a book recommendation personal site. I want to filter books by more than one criterion. For example, I want to display all books which are BOTH philosophy and science-fiction. Peter Watt's Blindsight being an example of this. Now I can do only a single criterion filter: I have a pivot table for Model Books public function genres() { return $this->belongsToMany('App\Genre', 'bookgenres', 'book_id', 'genre_id'); } Model Genre: public function books() { return $this-

What is local and remote side?

由 穿精又带淫゛_ 提交于 2019-12-10 03:26:35
问题 The questions below are related to database table relationships and the abstractions which SQLAlchemy provides for it. What is the difference between remote and local side? If there is remote_side then why not a local_side ? In the example given here how is parent_id " local " side? remote_side takes in a list so what are the elements of that list supposed to be? And if their are more then one elements then what exactly does that signify? I have read the docs several time but fail to

Parse Swift: User relations with a “friends request”

由 ぐ巨炮叔叔 提交于 2019-12-09 22:30:11
问题 I am kind of new to programming and parse, but I am really interested in it. I already know how to create a PFUser and how to set relations between Users, thanks to the tutorial provided on parse.com. The thing is they only have examples for one User following another User. I would prefer sending like a "Friends Request", for example like in instagram if a profile is not open to anybody. How can I code that? What do I need to think of? Example Code is very welcome :P 回答1: I would definitely

Laravel - How to setup a morphOne relationship

由 ⅰ亾dé卋堺 提交于 2019-12-09 18:58:52
问题 I'm trying to implement a morphable table for categories, right now I've the following. // Snippet Table - id - title - body // Post Table - id - title - body // Category Table - id - name I want to be able to morph the posts and snippets to have only one category, something like this: // Categorizable Table - category_id - categorizable_id - categorizable_type Do I have to have another model for this categorizable table? Or there is a way to set it without another model? So far I have this

PHP object parent/child recursion

由 放肆的年华 提交于 2019-12-09 18:11:33
问题 I've got a parent-child OO relationship . Parent obejcts has many child objects and every child object knows about it's parent by reference. The parent can be a child too ( basically its a tree ). When i do a var_dump() on the root object it says ["parent"]=> RECURSION many times and the generated description will be really long. I'm wondering if i do something wrong. If yes, i'm interested in the "best practice". Thanks for the help! 回答1: You're not doing anything wrong; you have a parent

Many-to-many self-referential relationship in sqlalchemy

由 微笑、不失礼 提交于 2019-12-09 14:47:31
问题 I'm trying to make a self-referential many-to-many relationship (it means that Line can have many parent lines and many child lines) in sqlalchemy like this: Base = declarative_base() class Association(Base): __tablename__ = 'association' prev_id = Column(Integer, ForeignKey('line.id'), primary_key=True) next_id = Column(Integer, ForeignKey('line.id'), primary_key=True) class Line(Base): __tablename__ = 'line' id = Column(Integer, primary_key = True) text = Column(Text) condition = Column

Neo4j Bidirectional Relationship

由 馋奶兔 提交于 2019-12-09 02:12:25
问题 Is there a way to create bidirectional relationship in Neo4j using Cypher? I would like the relationship to be bidirectional rather than making two unidirectional relationships in both directions For eg: (A)<-[FRIEND]->(B) Rather than: (A)-[FRIEND]->(B) (A)<-[FRIEND]-(B) Thanks in advance :) 回答1: No, there isn't. All relationships in neo4j have a direction, starting and ending at a given node. There are a small number of workarounds. Firstly, as you've suggested, we can either have two

Saving Entries to Database Relationship (many-to-many) Table from a Backbone Project

由 怎甘沉沦 提交于 2019-12-08 21:21:30
I currently have two Backbone models: user and project. I would like to have a Backbone view containing a form that enables the creation of a new project, and the association of currently existing users (entries from the database users table) with this project. When a user completes this form and clicks the save button, the new project should be saved into the database (in the projects table) and the relationship between the saved project and the related users should be saved into a relationship table (projects_users table, containing the corresponding project id and the user id for each

Ember 2, filter relationship models (hasMany, belongsTo) and calculate computed property based on relationships

由 梦想与她 提交于 2019-12-08 18:48:14
问题 These are my files: Models app/models/basket.js: export default DS.Model.extend({ name: DS.attr('string'), house: DS.belongsTo('house', { async: true }), boxes: DS.hasMany('box', { async: true }) }); app/models/box.js: export default DS.Model.extend({ qty: DS.attr('number'), basket: DS.belongsTo('basket'), cartLines: DS.hasMany('cart-line', { async: true }) }); app/models/cart-line.js: export default DS.Model.extend({ qty: DS.attr('number'), box: DS.belongsTo('box'), product: DS.belongsTo(
订阅 relationship