問題描述

我正在嘗試將圖像直接插入到 mysql 數據庫表中.在我的數據庫中，我總是得到 [BLOB - 0B].它不會將圖像插入表格中.我也沒有收到任何錯誤.我糊涂了..

I'm trying to insert the image into mysql database table directly. In my database I'm always getting [BLOB - 0B]. it doesn't insert images into table. I didn't get any error too. I'm confused..

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>

推薦答案

你差不多明白了，你想讓 PDO::PARAM_LOB 成為你上面創建的文件指針，而不是讀取的結果fp

You almost got it, you want PDO::PARAM_LOB to be a file pointer which you created above, not the result of reading the fp

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();
}

這篇關于pdo 直接將圖像插入數據庫 - 始終插入 BLOB - 0B的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！