0

我正在使用 sfGuardPlugin，尤其是 sfGuardGroup 表，它与“监视器”表（多对多表称为 ALERT）建立多对多关系：这是我的 schema.yml：

Monitor:
  tableName: monitor
  actAs:
    Timestampable: ~
  columns:
    id : {type: integer(4), primary: true, autoincrement: true}
    label: {type: string(45)}
    url: {type: string(80)}
    frequency: {type: integer}
    timeout: {type: integer}
    method: {type: enum, values: [GET, POST]}
    parameters: {type: string(255)}
    Groups:
      class: Groups
      local: monitor_id
      foreign: groups_id
      refClass: Alert

Groups:
  inheritance:
    extends: sfGuardGroup
    type: simple
  relations:
    Monitor:
      class: Groups
      local: id
      foreign: monitor_id
      refClass: Alert

Alert:
  actAs:
    Timestampable: ~
  columns:
    monitor_id: { type: integer(4), primary: true }
    groups_id: { type: integer, primary: true }
  relations:
    Monitor:
      foreignAlias: GroupMonitors
    sfGuardGroup:
      foreignAlias: GroupMonitors

在构建所有东西时，我得到了这个错误：

  SQLSTATE[42000]: Syntax error or access violation: 1072 Key column 'sf_guard_g
roup_id' doesn't exist in table. Failing Query: "CREATE TABLE alert (monitor_id
INT, groups_id BIGINT, created_at DATETIME NOT NULL, updated_at DATETIME NOT NUL
L, INDEX sf_guard_group_id_idx (sf_guard_group_id), INDEX id_idx (id), PRIMARY K
EY(monitor_id, groups_id)) ENGINE = INNODB". Failing Query: CREATE TABLE alert (
monitor_id INT, groups_id BIGINT, created_at DATETIME NOT NULL, updated_at DATET
IME NOT NULL, INDEX sf_guard_group_id_idx (sf_guard_group_id), INDEX id_idx (id)
, PRIMARY KEY(monitor_id, groups_id)) ENGINE = INNODB

有什么想法吗？

4

1 回答 1

2

在警报模型中 group_id 应该是 sf_guard_group_id 或者您可以使用 local_id 扩展 sfGuardGroup 的关系部分： group_id

Doctrine 试图猜测多对多表中使用的 id（如果你没有指定它们）。

于 2011-08-08T22:13:24.287 回答