更新数据库中的图像路径并在div中显示它们

问题描述：

这是我上传的图像代码。实际上，我可以上传和查看：更新数据库中的图像路径并在div中显示它们

$filename = $_FILES['upload_file']['name'].$SN; 
$target_path ="uploads/".basename($filename); 

if (move_uploaded_file($_FILES['upload_file']['tmp_name'], $target_path)) 
{ 
    echo "The file <b>".basename($filename)."</b> "."has been Uploaded.<br>"; 
    $result = $target_path ; 

    echo "Upload: " . $_FILES["upload_file"]["name"] . "<br>"; 
    echo "Type: " . $_FILES["upload_file"]["type"] . "<br>"; 
    echo "Size: " . ($_FILES["upload_file"]["size"]/1024) . " kB<br>"; 
    echo "Stored in: " . $_FILES["upload_file"]["tmp_name"]."<br/>"; 
    echo $result; 
    echo "<br/><img src='".$result."'"." alt=\"Image\" width=\"150px\" border=\"1\" />"; 
    echo "<form method=\"post\" name=\"myform\"><input type=\"checkbox\" name=\"agree_avatar\" onchange=\"myform.submit()\"></form>"; 

    if (!isset($_POST['agree_avatar'])) 
    { 
     echo "Select As Avatar"; 
    }

，直到我在这里可以看到echo "SELECT As Avatar"，但是当我检查的复选框就没有更新数据库，它给了我一个错误，There was an error uploading the file please try again!（$result是图片的路径，其被上传我想将它保存在数据库字段“头像”稍后调用它，只要我想）。
PS（我知道它缺乏安全检查，但我会后对工作）

if(isset($_POST['agree_avatar'])) 
    { 
     $sql = 'UPDATE register SET avatar="' . $result . '"' . ' WHERE SN="' . $SN . '"' . 'LIMIT 1'; 
     $run = mysql_query($sql); 

     if(!$run) 
     { 
      echo "DUH!<br/>"; 
     } 
     else 
     { 
      echo "Avatar selected "; 
     } 
} // end if from the beggining 
else 
{ 
echo "There was an error uploading the file please try again!"; 
}

答

我解决我自己的问题有一件事是$结果wasnot得到更新询问任何值，因此我改变我的脚本有点和结果我能够在数据库中增加值改变的脚本低于

 echo "<form method=\"GET\" name=\"myform\"><input type=\"checkbox\" name=\"agree\" onchange=\"myform.submit()\"><input type=\"hidden\" name=\"filename\" value=".$_FILES["upload_file"]['name'].$SN." ></form>"; 

     } 

     $filename = $_GET['filename']; 
     $target_path ="uploads/". basename($filename); 
     if(isset($_GET['agree'])) 
     { 

     $result = $target_path; 
     $sql = 'UPDATE register SET avatar="'.$result.'"'.' WHERE email="'.$user.'"'.' LIMIT 1'; 
     $run = mysql_query($sql) or die(mysql_errno()); 
      if($run) 
      { 
      echo "Avatar selected" ; 

      } 
     }

更新数据库中的图像路径并在div中显示它们

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