Question

我正在尝试直接将图像插入到mysql数据库表中。在我的数据库中，我总是得到[BLOB - 0B]。它不会将图像插入表格中。我也没有得到任何错误。我很困惑..

PHP

ini_set('display_startup_errors',1);
    ini_set('display_errors',1);
    error_reporting(-1);

    include('config.php');
     if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) 
      { 
          $tmpName  = $_FILES['image']['tmp_name'];  

          $fp = fopen($tmpName, 'r');
          $data = fread($fp, filesize($tmpName));
          $data = addslashes($data);
          fclose($fp);
      } 

      try
        {
            $stmt = $conn->prepare("INSERT INTO images ( picture ) VALUES ( '$data' )");
//          $stmt->bindParam(1, $data, PDO::PARAM_LOB);
            $conn->errorInfo();
            $stmt->execute();
        }
        catch(PDOException $e)
        {
            'Error : ' .$e->getMessage();
        }

HTML

<form action="upload.php" method="post">
<input id="image" name="image" type="file" />
<input type="submit" value="Upload" />
</form>

Answer 1

你几乎得到它，你希望PDO::PARAM_LOB成为你上面创建的文件指针，而不是读取fp的结果

if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) 
{ 
   $tmpName  = $_FILES['image']['tmp_name'];  

   $fp = fopen($tmpName, 'rb'); // read binary
} 

try
{
   $stmt = $conn->prepare("INSERT INTO images ( picture ) VALUES ( ? )");
   $stmt->bindParam(1, $fp, PDO::PARAM_LOB);
   $conn->errorInfo();
   $stmt->execute();
}
catch(PDOException $e)
{
   'Error : ' .$e->getMessage();
}

pdo直接将图像插入数据库 - 始终插入BLOB - 0B

1 个答案: