php - 困难的 SQL 查询语法

Question

大家好，这是我的 MySQL 查询，其中一个正在获取数据...

$videofetch = $conn->prepare("select * from user_followers as uf join videos as v on uf.followed_id = v.publisher_id where uf.follower_id = ? order by video_id desc limit 5");
$videofetch->execute(array(@$_SESSION ["userid"]));
$vid = $videofetch->fetchALL(PDO::FETCH_ASSOC); 

这段代码工作完美，但是当我尝试使用 AJAX 获取更多数据时，我无法编写正确的 sql 查询语法。

<?php
session_start();
if(isset($_POST["id"]) && !empty($_POST["id"])) {
  include('connectdb.php');
  $lastID = $_POST['id'];
  $videofetch = $conn->prepare("select * from  user_followers as uf join videos  as v  on uf.followed_id = v.publisher_id where uf.follower_id = ? order by video_id  desc limit 5");
  $videofetch->execute(array($_SESSION["userid"]));
  $vid = $videofetch->fetchALL(PDO::FETCH_ASSOC);
  ...

我要添加WHERE video_id < ".$lastID." ..我尝试了几种组合，但每次都显示语法错误。

注释:

1- 我正在从 AJAX 获取数据到 $lastID ;

2- $_SESSION ["userid"]处于事件状态，不用担心这个

3-SQL错误是:

Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'as v on uf.followed_id = v.publisher_id where uf.follower_id = '1' order by vid' at line 1 in C:\wamp64\www\hola.com\functions\getdatafoll.php on line 9

Answer 1

尝试将 mysql 关键字大写。阅读起来要好得多。

而且您已经有了一个 WHERE，因此只需将它与 AND 组合起来即可。

SELECT  *
    FROM  user_followers AS uf
    JOIN  videos AS v  ON uf.followed_id = v.publisher_id
    WHERE  uf.follower_id = ?
      AND  video_id < ".$lastID."
    ORDER BY  video_id DESC
    LIMIT  5 

但是更改'“.$lastID”。并附上一份准备好的声明。