dongxuxian6930 2011-11-04 12:08
浏览 54
已采纳

MySQL UPDATE无法正常工作

I have a bit of code which updates a table called job, but once the the page is executed it does not update the table. Here is the code:

$item = isset($_POST['item']);
$ref = isset($_POST['ref']);

$con = mysql_connect("$host","$username","$password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$db_name", $con);

$sql="UPDATE job SET item = '$item' WHERE ref='$ref'";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
header("location:index.php");

I have echoed out the $ref variable and it is there but it won't work if I put it in the WHERE clause.

  • 写回答

2条回答 默认 最新

  • dongzhunnai0140 2011-11-04 12:12
    关注
    $ref = isset($_POST['ref']);
    

    I have echoed out the $ref variable and it is there

    You aren't assigning the actual value of $_POST['ref'], you're only assigning whether or not it is set. Try:

    $ref = isset($_POST['ref']) ? mysql_real_escape_string($_POST['ref']) : NULL;
    

    You can check your query by reading the SQL string you've created: exit($sql)

    See also: What is SQL injection?

    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(1条)

报告相同问题?

悬赏问题

  • ¥15 求帮我调试一下freefem代码
  • ¥15 matlab代码解决,怎么运行
  • ¥15 R语言Rstudio突然无法启动
  • ¥15 关于#matlab#的问题:提取2个图像的变量作为另外一个图像像元的移动量,计算新的位置创建新的图像并提取第二个图像的变量到新的图像
  • ¥15 改算法,照着压缩包里边,参考其他代码封装的格式 写到main函数里
  • ¥15 用windows做服务的同志有吗
  • ¥60 求一个简单的网页(标签-安全|关键词-上传)
  • ¥35 lstm时间序列共享单车预测,loss值优化,参数优化算法
  • ¥15 Python中的request,如何使用ssr节点,通过代理requests网页。本人在泰国,需要用大陆ip才能玩网页游戏,合法合规。
  • ¥100 为什么这个恒流源电路不能恒流?