图片上传到数据库中？？？

给你几行主要的代码吧:if (is_uploaded_file($HTTP_POST_FILES['userfile']['tmp_name']))
{
    copy($HTTP_POST_FILES['userfile']['tmp_name'], "c:\tmp\temp_file");
}else{
    echo "Copy Error, File: $HTTP_POST_FILES['userfile']['name']";
}
      $imagename=$HTTP_POST_FILES['userfile']['name'];
      $type=$HTTP_POST_FILES['userfile']['type'];
      $size=$HTTP_POST_FILES['userfile']['size'];
      $mysqlImage = addslashes(fread(fopen"c:\tmp\temp_file", "rb"), $size));
      $link=mysql_connect("localhost","root","123456");
      mysql_select_db("image",$link) or die("Unable to select database");
      mysql_query("Insert Into image(image,image_size,FileType)
      VALUES('$imagename','$size','$type','$mysqlImage')",$link)
      or die("Query Error!");

解决方案 »

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echo "<td width=108><img src='showimage.php?id=".$id."' width=108 height=142></td>";

<?php
include_once('../includes/mysql.inc.php');
include_once('../includes/func.inc.php');
$temp_id=$_GET['id'];
$sql_win="select * from mag_list where mag_id=".$temp_id;
$res=$db->query($sql_win);
$row=$db->fetch_array($res);
if($row){ header("Content-Type:image/jpg");  //限于JPG
        echo $row['mag_cover'];
}?>放到数据库里似乎不是什么好的办法！
看成显示了，$imagename= addslashes(fread(fopen($imagename,"r"),filesize($imagename)));$sqlstr="Insert Into image(image,image_size,FileType) values(\"".$imagename."\",".$size.",\"".$type."\")";试试看
To：zhys9(OoP.plorer)
$mysqlImage = addslashes(fread(fopen"c:\tmp\temp_file", "rb"), $size))
这行少了一个（　　　我加了，运行后．出现：
Copy Error, File
Warning: fopen(c: mp emp_file) [function.fopen]: failed to open stream: Invalid argument in C:\Apache\htdocs\111.php on line 13Warning: fread(): supplied argument is not a valid stream resource in C:\Apache\htdocs\111.php on line 13
Query Error!
$uploaddir ='';
$uploadfile = $uploaddir.basename($_FILES['userfile']['name']);echo '<pre>';
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
   echo "File is valid, and was successfully uploaded.\n";
}
else {
   echo "Possible file upload attack!\n";
}echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";$imagename=$_FILES['userfile']['name'];
$size=$_FILES['userfile']['size'];
$type=$_FILES['userfile']['type'];$mylink=mysql_connect('localhost','root','123456');;
   mysql_select_db("image",$mylink);
$sqlstr="Insert Into image(image,image_size,FileType) values('".$imagename."','".$size."','".$type."')";
   $result=mysql_query($sqlstr) or die(mysql_errno().mysql_error());
  if ($result)
  echo $imagename."上传成功！<br/>";
{
    echo $imagename."上传失败！<br/>";}
看看什么错先
to:Robjuan()很感谢，没有什么问题．．．谢谢！
if ($result)
  echo $imagename."上传成功！<br/>";　　　／／这里加个else
{
    echo $imagename."上传失败！<br/>";}
if ($result)
  echo $imagename."上传成功！<br/>";
{
    echo $imagename."上传失败！<br/>";}不觉得这句话有问题么
各位高手：我现在要把图片从数据库中，显示到页面上．．为什么会只显示文件名：如4.jpg
怎样正常显示图片？
数据库建立为：
CREATE TABLE images (
  picid int(3) NOT NULL auto_increment,
  picdata longblob NOT NULL,
  pictext varchar(100) NOT NULL default '',
  PRIMARY KEY  (picid)
) TYPE=MyISAM;
<html>
<body>
<form action="add.php" method="post" enctype="multipart/form-data" name="form1">
  <input type="file" name="picfile[]">
  文件说明
  <input type="text" name="pictext[]">
  <br>
  <input type="file" name="picfile[]">
  文件说明
<input type="text" name="pictext[]">
  <br>
  <input type="file" name="picfile[]">
  文件说明
<input type="text" name="pictext[]">
  <br>
  <input type="file" name="picfile[]">
  文件说明
<input type="text" name="pictext[]">
  <br>
  <input type="submit" name="Submit" value="提交">
</form>
</body>
</html>add.php如下：if(trim($picfile[0])!=""){
    $link=@mysql_connect("localhost","root","root");
    //连接，用你的具体连接名替换root,具体连接密码替换root
        if($link==false)
           {
     echo "<script>alert('连接数据库时发生错误，请稍后再试!')</script>";
           }
    $res=mysql_select_db("kkk");
        if($res==false)
           {
     echo "<script>alert('打开数据库时发生错误，请稍后再试!')</script>";
           }
  $tmpset=0;
  // 存放成功加入的图片数目
  for($i=0;$i<=3;$i++)
      {
      // picfile[i]存放所提交的图片信息(文件路径)
      if(trim($picfile[$i])!="")
         {
         //读取数据
         $fp=fopen($picfile[$i],"r");
         $picdata=fread($fp,filesize($picfile[$i]));
         fclose($fp);
         //加上必要的标志符号
         $picdata=addslashes($picdata);
         //用具体的数据表名代替images
         //pictext[i]存放所提交的图片的文字说明
         $qu="insert into images(picdata,pictext) values('$picdata','$pictext[$i]')";
         $res=@mysql_query($qu,$link);
               if($res==false)
                   {
       echo "<script>alert('图片 ".$i." 提交失败!')</script>";
                   continue;
                   }
         $tmpset=$tmpset+1;
         }
       }
  echo "<script>alert('操作成功!实际入库图片数 ".$tmpset." 张')</script>";
}
?>程序运行没有错误．但存不进数据库中．．请问各位高手怎么解决．．．．还有下面是图片显示出来程序．．．请问能否行得通．．．
$link=@mysql_connect("localhost","root","root");
mysql_select_db("kkk");
$qu="select picid,picdata from images where picid=$picid";
$res=@mysql_query($qu,$link);
$num=mysql_num_rows($res);
   if($num==0)
      {
      print "<br><br><br>";
      print "<p><b>没有这张图片!</b></p>";
      exit();
      }
$row=@mysql_fetch_row($res);
header("Content-type:image/");
echo $row[1];
to :ShadowSniper(青春不等人呀...)
可以，请给我个例子．．．．现在就要用啊．．．呵．．．．．．．