Question

我有三张桌子

table_users: id_user | name
      table_posts: id_post | id_user | post
            table_follows: id_follow | id_follower | id_followed

第一个表显示了用户，第二个表显示了他们发布的帖子＆amp;第三，用户之间的关系（跟随者/跟随者）

因此，例如，如果我有用户Mattew id=3，我想要检索他所关注的所有人“发布”的帖子。在这种情况下，亚当id=1，汤姆id=7＆amp; Zoe id=9之后是Mattew。

我已经开发了一个有效的php/mysql代码，但我想仅使用一个SQL语句来改进它。

//retrieve all the people followed by Mattew
$followeds=mysql_query("SELECT * FROM table_follows WHERE id_follower='3' /*Mattew*/");

//create an array with the ids
$array_followeds=array();

while($fwd=mysql_fetch_array($followeds)){
    $id_followed=$fwd['id_followed'];
    $array_followeds[]="id_user='".$id_followed."'";
}

//if exists people followed by Mattew
if(count($array_followeds)>0){
    $array_followeds=implode(' OR ', $array_followeds);
}else{/*$array_followeds="id_user=0";*/}

//main query: shows all the posts of the people Mattew follows
$main_query=mysql_query("SELECT * FROM $table_posts WHERE ($array_followeds) AND id_user!='3'/*Mattew*/");

while($posts=mysql_fetch_array($main_query)){
      /*results */
}

我想建议如何使用单个语句改进Sql查询并避免使用数组。

Answer 1

选择您需要的字段

defmodule Chatroom.LobbyChannel do
  require Runner
  use Phoenix.Channel

  {:ok, pid: spawn(fn -> Runner.input() end)}

  avatar = %{x: 0, y: 0}

  def join("lobby", _payload, socket) do
    {:ok, socket}
  end

  def handle_in("new_message", payload, socket) do
    case payload["message"] do
      "hello" -> broadcast! socket, "new_message", payload
      "jump" -> send pid, {:jump, self, [avatar: avatar, socket: socket]}
      _ -> nil
    end

    {:noreply, socket}
  end

  def handle_reply do
    receive do
        {:result, mover, socket} -> broadcast! socket, "new_message", [name: "State", message: "5"]
    end
  end
end

返回用户Mysql后面的所有帖子

1 个答案: