php - 通过表单 Cookie 和 GET

Question

我正在使用 $_GET 函数在整个表单中传递数据库字段。但是，当我尝试从数据库中提取信息时遇到了一个绊脚石:

<?php
$prodname=$_GET["q"];
?>

 <h3>Product Name: <u><?php echo $prodname; ?></u></h3><br />
  <?php
$con = mysql_connect("localhost","cl49-vogalcms","vogalcms");
if (!$con) 
  {
   die('Could not connect: ' . mysql_error());
   }

  $prodname=$row['prodname'];
 $catagory=$row['catagory'];

@mysql_select_db("cl49-XXX",$con)or die('Unable to select database ln 60:'.mysql_error());
$result=mysql_query("SELECT * FROM products WHERE prodname=$prodname")or die('ln 61 :'.mysql_error());
$cnt=$_COOKIE["count"];


setcookie("user",$myid,time()+10000);
mysql_close($con);
?>

  <form name="newad" method="post" enctype="multipart/form-data"  action="drtsavepic.php?q=<?php echo"$prodname"; ?>">  <br> <Br>
   <input type="file" name="image">
<input name="Submit" type="submit" value="Upload image">

  </form>

当我运行这段代码时，出现以下错误

ln 61 :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

谁能阐明这个问题？

Answer 1

尝试替换此 mysql_query("SELECT * FROM products WHERE prodname=$prodname")

由

mysql_query("SELECT * FROM products WHERE prodname='".$prodname."'")