圖片/文件上傳到mysql數據庫不起作用

Question

我目前有一段時間上傳圖像到我的數據庫。我目前有多個變量/輸入從一個表單上傳 - 如果這些輸入是圖像文件上傳的話。該文件似乎使它做數據庫，但是當我嘗試通過PHP腳本檢索圖像時，它只是返回「Array」，而不是圖像。任何幫助？謝謝！

這裏的上傳代碼：

   // if the form's submit button is clicked, we need to process the form 
      if (isset($_POST['submit'])) 
      { 
        // get the form data 
          $projectname = htmlentities($_POST['projectname'], ENT_QUOTES); 
          $item = htmlentities($_POST['item'], ENT_QUOTES); 
          $description = htmlentities($_POST['description'], ENT_QUOTES); 
          $neededby = htmlentities($_POST['neededby'], ENT_QUOTES); 
          $shipping= htmlentities($_POST['shipping'], ENT_QUOTES); 
          $revisions = htmlentities($_POST['revisions'], ENT_QUOTES); 
          $price = htmlentities($_POST['price'], ENT_QUOTES); 
          $paid = htmlentities($_POST['paid'], ENT_QUOTES); 
          $ordered1 = htmlentities($_POST['ordered1'], ENT_QUOTES); 
          $ordered2 = htmlentities($_POST['ordered2'], ENT_QUOTES); 
          $ordered3 = htmlentities($_POST['ordered3'], ENT_QUOTES); 
          $received1 = htmlentities($_POST['received1'], ENT_QUOTES); 
          $received2 = htmlentities($_POST['received2'], ENT_QUOTES); 
          $received3 = htmlentities($_POST['received3'], ENT_QUOTES); 
          $shipped1 = htmlentities($_POST['shipped1'], ENT_QUOTES); 
          $shipped2 = htmlentities($_POST['shipped2'], ENT_QUOTES); 
          $shipped3 = htmlentities($_POST['shipped3'], ENT_QUOTES); 
          $tracking = htmlentities($_POST['tracking'], ENT_QUOTES); 
          $delivered = htmlentities($_POST['delivered'], ENT_QUOTES); 
          $thestatus = htmlentities($_POST['thestatus'], ENT_QUOTES); 
          $photo=($_FILES['photo']); 




        if ($projectname == '') 
          { 
            // if they are empty, show an error message and display the form 
            $error = 'ERROR: Please fill in project name!'; 
            renderForm($projectname, $item, $description, $neededby, $shipping, $revisions, $price, $paid, $ordered1, $ordered2, $ordered3, $received1, $received2, $received3, $shipped1, $shipped2, $shipped3, $tracking, $delivered, $thestatus, $photo, $error, $id); 
          } 

          else 
          { 
          // insert the new record into the database 
          if ($stmt = $mysqli->prepare("INSERT todo (projectname, item, description, neededby, shipping, revisions, price, paid, ordered1, ordered2, ordered3, received1, received2, received3, shipped1, shipped2, shipped3, tracking, delivered, photo, thestatus) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)")) 
          { 
            $stmt->bind_param("sssssssssssssssssssss", $projectname, $item, $description, $neededby, $shipping, $revisions, $price, $paid, $ordered1, $ordered2, $ordered3, $received1, $received2, $received3, $shipped1, $shipped2, $shipped3, $tracking, $delivered, $photo, $thestatus); 
            $stmt->execute(); 
            $stmt->close(); 
          } 

          // show an error if the query has an error 
          else 
          { 
            echo "ERROR: Could not prepare SQL statement."; 
          } 



          // redirec the user 
          header("Location: main.php"); 
        } 

      } 

和文件檢索代碼：

<?php 
mysql_connect("localhost","MYUSER","MYPASS"); 
mysql_select_db("MYDB"); 
$query = "SELECT photo FROM todo where id=$id"; 
$result = MYSQL_QUERY($query); 
$data = MYSQL_RESULT($result,0,"photo"); 
Header("Content-type: $type"); 
print $data; 
?> 

MySQL的列是一個BLOB類型。

這裏是一個形象，你可以上我說的一些視覺效果： http://i.imgur.com/DYHHx.png

Answer 1

試試這個教程

http://www.tizag.com/phpT/fileupload.php

把文件名可變然後將其插入到數據庫中（我期待你知道如何從數據庫檢索數據）

Answer 2

$fileName = $_FILES['image']['name']; 
$tmpName = $_FILES['image']['tmp_name']; 
$fileSize = $_FILES['image']['size']; 
$fileType = $_FILES['image']['type']; 

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

if(!get_magic_quotes_gpc()) 
{ 
    $fileName = addslashes($fileName); 
} 


//and here your insert query as i remember you can try it 


HTML CODE: 

    <input type=\"file\" name=\"image\" /> 


and here is how you retrive it 

echo '<img src="data:image/jpeg;base64,' . base64_encode($row['imageContent']) . '" />'; 

但我不建議你這樣做，因爲它會使你的數據庫沒有加載快，所以保存圖像以一個文件夾，onlyt在DATABSE

記我已得到了一個論壇和唐該代碼的名稱;噸記得它的名字對不起

Answer 3

這通常是一個非常糟糕的主意，圖像直接上傳到數據庫。原因是因爲過了一段時間，從數據庫讀取和上傳文件會使數據庫過載。

更好的解決方案是將圖像上傳到服務器上的文件夾，然後將文件名和位置保存在數據庫中。