Question

我想在这里做两件事，第一件工作正常（它将数据从一个表复制到另一个表）。然而，第二部分不是......这部分应该将值1添加到列＆＃34;得分＆＃34;但只有最新的结果（由排序＆＃34; date_played＆＃34;定义），目前我正在做的是向表中添加一个新的结果..

非常感谢。

<?php
header("Cache-Control: no-cache");
date_default_timezone_set('Europe/London');

// open DB connection
require_once("DbConnect.php");

// fetch now playing song and copy ratings table
$result = $db->query(
    "INSERT INTO rating(`artist`, `title`, `label`, `albumyear`, 
                        `date_played`, `duration`, `picture`, 
                        `filename`, `songid`)" .
     "SELECT `artist`, `title`, `label`, `albumyear`, `date_played`, 
             `duration`, `picture`, `filename`, `songid` " .
     "FROM historylist  
      ORDER BY `date_played` DESC 
      LIMIT 1 ");


$score  = $db->query(
    "INSERT INTO rating (`score`)".
    "VALUES ('1')".
    "ORDER BY `date_played` DESC LIMIT 1 ");

?>

Answer 1

UPDATE rating set `score`=`score`+1 ORDER BY `date_played` DESC LIMIT 1

<强> 修改

更好的方法是获取最后插入的ID并使用它来定位行

$last_id = $db->insert_id; // assuming mysqli API

$db->query("UPDATE rating SET `score` = `score` + 1 WHERE ID=$last_id");

使用PHP＆amp; amp;集成两个数据库查询MYSQL

1 个答案: