2015-03-19 70 views
-2

我有一个表格,其中我必须上传图像的类别,并将选定的类别图像移动到特定的文件夹,但是当我提交表单时发生以下错误 (!)通知:未定义的索引:图片在C:\ wamp \ www \ movies海报admin \ insertform.php在线26 图片是文件的名称,但不知道为什么它没有得到值 我知道这是一个很基本的问题,但我不知道为什么这个错误就要上传图像,但未定义索引的错误

 $imageprice=$_POST['imageprice']; 
     $imagedescription=$_POST['imagedescription']; 
     $type=$_POST['category']; 

     if ($type="celeb") { 
      $uploadDir = 'C:/wamp/www/Movies Poster Site/celeb/'; //Image Upload Folder 
      $fileName = $_FILES['pic']['name']; 
      $tmpName = $_FILES['pic']['tmp_name']; 
      $fileSize = $_FILES['pic']['size']; 
      $fileType = $_FILES['pic']['type']; 
      $filePath = $uploadDir . $fileName; 
      $filename = $fileName; 
      $ext = pathinfo($filename, PATHINFO_EXTENSION); 
      //$date = date_create(); 
      //$fileName= date_timestamp_get($date).".".$ext; 
      $filePath = $uploadDir . $fileName; 
      if (isset($fileName)) { 

       # code... 

      $result = move_uploaded_file($tmpName, $filePath); 
      } 
      if(!get_magic_quotes_gpc()) 
      { 
       $fileName = addslashes($fileName); 
       $filePath = addslashes($filePath); 
      } 

`

`$sql="INSERT INTO celeb ". 
     "(image_name,price,description,Type) ". 
     "VALUES ". 
     "('$filePath','$imageprice','$imagedescription','$type') "; 
     mysql_select_db('poster'); 
     $retval = mysql_query($sql, $connection); 
     if(! $retval) 
     { 
      die('Could not enter data: ' . mysql_error()); 
     } 
     else 
     { 
     echo" <div class='end'>"; 
     echo "Data Entered"; 
     echo"</div>"; 
     mysql_close($connection); 
     } 

这里是形式

<form method="post"> 
<div class="insertform"> 

<!--   <input type="text" placeholder="image.jpg" name="imagename" id="imagename"><br> 
--> 

Select Image Category:<br> 
<input type="radio" name="category" value="comic">Comic Posters<br> 
<input type="radio" name="category" value="celeb">Celeb Posters<br> 
<input type="radio" name="category" value="islamic">Islamic Posters<br> 
<input type="radio" name="category" value="tv">TV Posters<br> 
<input type="radio" name="category" value="motor">Motor Posters<br /> 
<input type="radio" name="category" value="sports">Sports Posters<br><br /> 

Image Name :<br> 
<input style="margin-left:10px;" id="uploadImage" required="required" name="pic" type="file" size="4 
       000000" accept="image/gif, image/jpeg, image/x-ms-bmp, image/x-png" > <br> 
    Image Price :<br> 
     <input type="text" placeholder="Rs xxx" name="imageprice" id="price"><br> 
Image Description :<br> 
     <input type="text" name="imagedescription" id="description" height="300px"><br /><br> 


<input type="submit" name="submit" value="Submit"> 

</div> 
+0

添加以下PHP表单看起来不错。 – 2015-03-19 07:13:06

+0

切线,但在你的'if'中你需要'==',而不是'='。 – deceze 2015-03-19 07:13:55

+0

您需要将'enctype ='multipart/form-data''作为参数之一添加到您的表单中。否则表单将不会发布文件/图像数据。 – 2015-03-19 07:18:10

回答

-1

尝试这样,

$fileName = $_FILES['pic']['name'][0];