php - 尝试更新小时表中的 time_out 时出现语法错误

Question

我试图将 time_out 时间更新为捕获用户 time_in 的最后一个 id，我的以下代码给了我这个错误

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 '' at line 1

这是我的 logout.php

session_start();
if(!isset($_SESSION['MEMBER_ID']) || (trim($_SESSION['MEMBER_ID']) == '')) {
        header("location: index.php");
        exit();
}   

date_default_timezone_set('America/New_York');
$dt = date("Y-m-d h:i:s");      
$id = $_SESSION['MEMBER_ID'];
$sql = "UPDATE hours SET time_out = '".$dt."' WHERE member_id IN (SELECT MAX(hours_id) FROM hours WHERE member_id = '".$id."'";
mysql_query($sql) or die(mysql_error());

    //Unset the variables stored in session
    unset($_SESSION['MEMBER_ID']);
    unset($_SESSION['LOGIN_NAME']);
    unset($_SESSION['PASS']);

这是架构

 `hours_id` int(11) NOT NULL AUTO_INCREMENT,
  `member_id` int(11) NOT NULL,
  `member_name` varchar(32) NOT NULL,
  `team` varchar(32) NOT NULL,
  `time_in` datetime NOT NULL,
  `time_out` datetime DEFAULT NULL,
  `dated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
    unset($_SESSION['TEAM']);   

某人的这个查询可能出了什么问题？

Answer 1

像这样关闭查询中的圆括号吗？

$sql = "UPDATE hours SET time_out = '".$dt."' WHERE member_id IN (SELECT MAX(hours_id) FROM hours WHERE member_id = '".$id."')";

注意:停止使用已弃用的 mysql_* 版本。有更好的选择，例如 mysqli 或 pdo

编辑

通过单个查询，您可以像这样解决问题

$sql = "UPDATE hours SET time_out = '".$dt."' WHERE member_id IN (SELECT MAX(hours_id) FROM (SELECT MAX(hours_id) FROM hours WHERE member_id = '".$id."') AS t)";

请告诉我它是否适合您