上傳和顯示圖像php mysqli

我需要幫助我的顯示圖像，顯示是正確的，但我不知道爲什麼圖像不會出現。上傳和顯示圖像php mysqli

CREATE TABLE `images` (
    `id` int(11) NOT NULL, 
    `name` varchar(100) DEFAULT NULL, 
    `size` int(11) DEFAULT NULL, 
    `type` varchar(20) DEFAULT NULL, 
    `content` mediumblob 
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

，這是我的「upload.php的」

<?php 

// Check for post data. 
if ($_POST && !empty($_FILES)) { 
    $formOk = true; 

    //Assign Variables 
    $path = $_FILES['image']['tmp_name']; 
    $name = $_FILES['image']['name']; 
    $size = $_FILES['image']['size']; 
    $type = $_FILES['image']['type']; 

    if ($_FILES['image']['error'] || !is_uploaded_file($path)) { 
     $formOk = false; 
     echo "Error: Error in uploading file. Please try again."; 
    } 

    //check file extension 
    if ($formOk && !in_array($type, array('image/png', 'image/x-png', 'image/jpeg', 'image/pjpeg', 'image/gif'))) { 
     $formOk = false; 
     echo "Error: Unsupported file extension. Supported extensions are JPG/PNG."; 
    } 
    // check for file size. 
    if ($formOk && filesize($path) > 500000) { 
     $formOk = false; 
     echo "Error: File size must be less than 500 KB."; 
    } 

    if ($formOk) { 
     // read file contents 
     $content = file_get_contents($path); 

     //connect to mysql database 
     if ($conn = mysqli_connect('localhost', 'root', '', 'upload')) { 
      $content = mysqli_real_escape_string($conn, $content); 
      $sql = "insert into images (name, size, type, content) values ('{$name}', '{$size}', '{$type}', '{$content}')"; 

      if (mysqli_query($conn, $sql)) { 
       $uploadOk = true; 
       $imageId = mysqli_insert_id($conn); 
      } else { 
       echo "Error: Could not save the data to mysql database. Please try again."; 
      } 

      mysqli_close($conn); 
     } else { 
      echo "Error: Could not connect to mysql database. Please try again."; 
     } 
    } 
} 
?> 

<html> 
    <head> 
     <title>Upload image to mysql database.</title> 
     <style type="text/css"> 
      img{ 
       margin: .2em; 
       border: 1px solid #555; 
       padding: .2em; 
       vertical-align: top; 
      } 
     </style> 
    </head> 
    <body> 
     <?php if (!empty($uploadOk)): ?> 
      <div> 
       <h3>Image Uploaded:</h3> 
      </div> 
      <div> 
       <img src="image.php?id=<?=$imageId ?>" width="150px"> 
       <strong>Embed</strong>: <input size="25" value='<img src="image.php?id=<?=$imageId ?>">'> 
      </div> 

      <hr> 
     <?php endif; ?> 

     <form action="<?=$_SERVER['PHP_SELF']?>" method="post" enctype="multipart/form-data" > 
      <div> 
      <h3>Image Upload:</h3> 
      </div> 
      <div> 
      <label>Image</label> 
      <input type="hidden" name="MAX_FILE_SIZE" value="500000"> 
      <input type="file" name="image" /> 
      <input name="submit" type="submit" value="Upload"> 
      </div> 
     </form> 
    </body> 
</html>

，那麼這是我的「image.php」

<?php 

    // verify request id. 
    if (empty($_GET['id']) || !is_numeric($_GET['id'])) { 
     echo 'A valid image file id is required to display the image file.'; 
     exit; 
    } 

    $imageId = $_GET['id']; 

    //connect to mysql database 
    if ($conn = mysqli_connect('localhost', 'root', '', 'upload')) { 
     $content = mysqli_real_escape_string($conn, $content); 
     $sql = "SELECT type, content FROM images where id = {$imageId}"; 

     if ($rs = mysqli_query($conn, $sql)) { 
      $imageData = mysqli_fetch_array($rs, MYSQLI_ASSOC); 
      mysqli_free_result($rs); 
     } else { 
      echo "Error: Could not get data from mysql database. Please try again."; 
     } 
     //close mysqli connection 
     mysqli_close($conn); 

    } else { 
     echo "Error: Could not connect to mysql database. Please try again."; 
    } 

    if (!empty($imageData)) { 
     // show the image. 
     header("Content-type: {$imageData['type']}"); 
     echo $imageData['content']; 
    } 
?>

每次我嘗試上傳圖片過程都很成功，但我上傳的圖片沒有出現。但那隻會出現圖像錯誤。請任何人可以幫助我？

或者有沒有人可以給我一個教程如何使用數據庫上傳和顯示圖像？

來源

2016-03-30 Adam Projo

-1

你必須提取['content']一段時間。就像這樣：

$rs = mysqli_query($conn, $sql); 
    while ($imageData = mysqli_fetch_assoc($rs)) { 
     $content = $imageData['content']; 
    } 
    header("Content-type: $content "); 
    echo $content;

我會建議你檢查和更改：

$sql = "SELECT type, content FROM images where id = {$imageId}";

爲

$sql = "SELECT type, content FROM images where id = '$imageId' ";

來源

2016-03-30 06:08:44

OWH OKE。我會在下次嘗試。，因爲它似乎我已經解決了與另一個腳本。但感謝您的幫助。它幫助我。 –

上傳和顯示圖像php mysqli

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