2014-03-26 14 views
0

我是PHP新手,所以這可能很簡單。 我只想在圖像上傳並通過POST發送到腳本後顯示圖像。但我無法看到圖像。php imagecreatefromjpeg無法正常工作

請找到下面的代碼: php文件的名稱是test.php的

<?php 

if ($_SERVER['REQUEST_METHOD'] == 'POST'){ 

    if(isset($_FILES['photo']) 
    && is_uploaded_file($_FILES['photo']['tmp_name']) 
    && $_FILES['photo']['error'] == UPLOAD_ERR_OK){ 

     if ($_FILES['photo']['type']=='image/jpeg') { 

      $tmp_img = $_FILES['photo']['tmp_name']; 
      $image = imagecreatefromjpeg($tmp_img); 
      header('Content-Type: image/jpeg'); 
      imagejpeg($image,'',90); 
      imagedestroy($image); 

     }else{ 
      echo "Uploaded file was not JPEG","</br>"; 
     } 
    }else{ 
     echo "No file uploaded","</br>"; 

    } 
}else{ 
    ?> 
    <form action="test.php" method="post" enctype="multipart/form-data"> 
     <label for="photo">Photo : </label> 
     <input type="file" name="photo"/> 
     <input type="submit" value="Upload a JPEG photo"/> 
    </form> 

<?php } ?> 
+0

你有['PHP GD'(HTTP:// WWW。 php.net/manual/es/book.image.php)安裝? –

回答

0

你需要NULL路過這裏:

 imagejpeg($image,NULL,90); 

從文檔下Parameters

filename 

    The path to save the file to. If not set or NULL, the raw image stream will be outputted directly. 

    To skip this argument in order to provide the quality parameter, use NULL.