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我有兩個表A(Utenti)e B(Dispositivi);在B上有一個A的外鍵(一個Utente-> many Dispositivi)。我使用了symfony管理生成器。我可以爲每個Utente生成一個鏈接,它列出Dispositivi視圖中的所有相關Dispositivi。這可能嗎?symfony外鍵一對多
的schema.yml
Dispositivi:
connection: doctrine
tableName: dispositivi
columns:
id_dispositivo:
type: integer(4)
fixed: false
unsigned: false
primary: true
autoincrement: true
device_id:
type: string(255)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
tipo:
type: string(255)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
utente_fk:
type: integer(4)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
relations:
Utenti:
local: utente_fk
foreign: id_utente
type: one
Utenti:
connection: doctrine
tableName: utenti
columns:
id_utente:
type: integer(4)
fixed: false
unsigned: false
primary: true
autoincrement: true
username:
type: string(255)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
password:
type: string(255)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
tipo:
type: string(255)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
relations:
Dispositivi:
local: id_utente
foreign: utente_fk
type: many
Servizi:
local: id_utente
foreign: utente_fk
type: many
我這樣做。我有相關的dispositivis foregn鍵。我真正的問題是我無法列出所有Utente相關Dispositivis。我嘗試dispositivo/list?id_dispositivo = 1,但不起作用。換句話說,我想在生成的後端的dispositivis視圖中通過它們的id來過濾dispositivis。我希望你明白我在說什麼 – Omega 2011-05-24 08:43:44
你將不得不看看lib/filters並在那裏做出改變。這可能有所幫助:http://www.prettyscripts.com/framework/symfony/symfony-customize-admin-generator-filter – Kevin 2011-05-25 02:52:30