Question

我有两张桌子：Conversations和ConversationParticipants。我需要获取用户1和2都参与的对话列表。在MySQL中，查询将是：

SELECT conversation_participants.conversation_id FROM conversation_participants
JOIN conversations t1
    ON t1.conversation_id = conversation_participants.conversation_id
    AND conversation_participants.user_id = 11
JOIN conversation_participants t2
    ON t1.conversation_id = t2.conversation_id
    AND t2.user_id = 2

然而，在Sequelize中我无法理解如何设置模型关系以便我可以进行单个查询。我尝试了这个没有成功（请注意，这几乎是伪代码，为了清楚起见这样报告）：

var Conversation = sequelize.define('conversations', {
    conversationId:  Sequelize.INTEGER.UNSIGNED,
});

var ConversationParticipant = sequelize.define('conversation_participants', {
    participationId:  Sequelize.INTEGER.UNSIGNED,
    userId:  Sequelize.INTEGER.UNSIGNED,
    conversationId : Sequelize.INTEGER.UNSIGNED,        
});
Conversation.hasMany(ConversationParticipant, { as : 'Participants'});

然后

ConversationParticipant.findAll({
    where : { userId : 1 },
    include : [{
        model : Conversation,
        as : 'conversation',
        include : [{
            model : ConversationParticipant,
            as : 'Participants',
            where : { userId : 2 }
        }]
}]

我收到以下错误： Error: conversation_participants is not associated with conversations!。有什么想法吗？

Answer 1

您可以采用的一种方法是找到参与者拥有用户1或2的对话，然后过滤包含这两者的对话：

<g id="Artwork_18" class="cerebellum">
  <defs><path id="SVGID_1_" d="M1097.44 638.07l4.77-53.1 53.5 4.8-4.77 53.1z"/></defs>
  <clipPath id="SVGID_2_">
    <use xlink:href="#SVGID_1_" overflow="visible"/>
  </clipPath>
  <path d="M1114.7 586l-13 5.2-1.9 21.2z" class="st1 poly"/> <!-- ... more paths... --> </g>

Answer 2

您缺少ConversationParticipant定义中的belongsTo关联。 Sequelize模型需要显式关联，因为您尝试通过Conversation实例访问ConversationParticipant，即使关联Conversation不是ConversationParticipant。

除了hasMany关联之外还有这样的东西：

ConversationParticipant.belongsTo(Conversation, { as : 'conversation',
  foreignKey: 'conversationId', targetKey: 'conversationId'});

然后从conversationId模型中移除ConversationParticipant字段定义，因为belongsTo也会为您创建。{/ p>

Sequelize在两个表上连接两次

2 个答案: