對(duì)不起，如果這是一個(gè)愚蠢的問題，但我是 Laravel 的新手。我有兩個(gè)模型和一個(gè)數(shù)據(jù)透視表：用戶id | name | passwordpublic function conversations(): ?BelongsToMany{  return $this->belongsToMany(Conversation::class)->withTimestamps();}對(duì)話idpublic function users(): ?BelongsToMany{  return $this->belongsToMany(User::class)->withTimestamps();}對(duì)話用戶id | conversation_id | user_id我創(chuàng)建一個(gè)對(duì)話并為用戶分配同步，如下所示：$user->conversations()->syncWithoutDetaching($conversation);$targetUser->conversations()->syncWithoutDetaching($conversation);用戶可以有很多對(duì)話，對(duì)話可以有多個(gè)用戶。這很好，但是當(dāng)我想與兩個(gè)特定用戶進(jìn)行對(duì)話時(shí)，我不知道利用 ORM 來找到他們各自所在的對(duì)話的最佳方式。我目前正在使用下一種方法，該方法有效，但感覺使用 ORM 有更好的方法：/** * Get a conversation by a target user id. * * @param int $targetUserId * @return mixed */public function getConversationByTargetUserId(int $targetUserId){    // Get the current user.    $user = Auth::guard()->user();    // Check the user exists.    if (!$user) {        throw new HttpException(500);    }    /**     * Get all pivot tables where the     * user ID is from the current user.     */    $userConversationIdsArray = DB::table('conversation_user')->where('user_id', $user->id)->pluck('conversation_id');    /**     * Get all pivot tables where the user     * id is equal to the target id, and is     * also owned by the current user. Return     * the first instance that we come across.     */    $targetConversation = DB::table('conversation_user')->where(['conversation_id' => $userConversationIdsArray, 'user_id' => $targetUserId])->first();    /**     * Return the conversation.     */    return Conversation::find($targetConversation->conversation_id);}感謝您的時(shí)間 ：）