Question

我正在使用sfGuardPlugin，特别是与“监视器”表（多对多表称为ALERT）的多对多关系中使用的sfGuardGroup表：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

当构建所有thigs时，我收到此错误：

  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

有任何想法吗？

Answer 1

在Alert模型中，group_id应为sf_guard_group_id 要么或者，您可以使用local_id：group_id

扩展sfGuardGroup的关系部分

Doctrine试图猜测多对多表中使用的id（如果你没有指定它们）。

Symfony 1.4 / doctrine / sfGuard schema.yml在多对多关系中使用sfDoctringGuardPlugin的sfGuardGroup表

1 个答案: