这是我拥有的代码,但是当我试图从url获取变量时会出现此错误:警告:mysql_fetch_array():提供的参数不是有效的mysql结果资源
URL变量DEVID是一长串字符、数字、破折号和下划线。有什么问题吗?

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

mysql_select_db("database", $con);

$result = mysql_query("SELECT * FROM $user WHERE DEVID=$DEVID");


while($row = mysql_fetch_array($result))
  {
  if (($row["FN"]) == NULL){
echo '<meta http-equiv="refresh" content="1;url=../register/default.php?user=';
echo $_GET["user"];
echo '&DEVID=';
echo $_GET["DEVID"];
echo '">Please hold, we are taking you to the registration page.<br/><br/>';
}

  }


mysql_close($con);
?>

最佳答案

如果$DEVID是一个VARCHAR字段,那么在SQL查询中需要在它周围加上单引号:

SELECT * FROM $user WHERE DEVID='$DEVID'

10-07 17:31