dsb238100 2013-05-23 02:55
浏览 9
已采纳

PHP mySQL查询使用变量更新表中的行

My PHP file contains the following function. It works when I set the review column to '$review' and the IdUser to 2. But I need to have the IdUser set to the variable $user. What is the correct syntax to set IdUser to the variable instead of a constant? (preferably in a way that avoids SQL injection attacks).

function addRatings2($review, $user) {  
    //try to insert a new row in the "ratings" table with the given UserID
    $result = query("UPDATE ratings SET review ='$review' WHERE IdUser = 2 order by dateTime desc limit 1");    
}
  • 写回答

4条回答 默认 最新

  • doupeizheng3918 2013-05-23 03:02
    关注

    Hi the right syntax is to use

    {$var} wherever you want the current value of var to appear, so in your case it would be

    $result = query("UPDATE ratings SET review ='{$review}' WHERE IdUser = {$user}
    order by dateTime desc limit 1");
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(3条)

报告相同问题?

悬赏问题

  • ¥15 求lingo代码和思路
  • ¥15 公交车和无人机协同运输
  • ¥15 stm32代码移植没反应
  • ¥15 matlab基于pde算法图像修复,为什么只能对示例图像有效
  • ¥100 连续两帧图像高速减法
  • ¥15 如何绘制动力学系统的相图
  • ¥15 对接wps接口实现获取元数据
  • ¥20 给自己本科IT专业毕业的妹m找个实习工作
  • ¥15 用友U8:向一个无法连接的网络尝试了一个套接字操作,如何解决?
  • ¥30 我的代码按理说完成了模型的搭建、训练、验证测试等工作(标签-网络|关键词-变化检测)