Php 保存多个附件并检索它们

我有一个简单的html表单，它发送一封电子邮件，带有多个附件（最多5个）。我正在保存电子邮件，地址为phpadmin。现在我还想保存多个附件。 如何保存多个附件，然后在另一页上重试

我使用phpmailer发送电子邮件和附件，发送附件的循环是：

  $MyUploads = array();
  foreach(array_keys($_FILES['attach']['name']) as $key) 
{  
$filesize = $_FILES['attach']['size'][$key];
$extention = pathinfo ($_FILES['attach']['name'][$key] ,PATHINFO_EXTENSION );
$name=$_FILES['attach']['name'][$key];
    $filePath = $Location . $name;
    $tmpName = $_FILES['attach']['tmp_name'][$key];


if(move_uploaded_file($tmpName, $filePath))
    {$MyUploads[] = $Location;}
     else
    {$MyUploads[]='';}
     print_r($MyUploads);


$source = $_FILES['attach']['tmp_name'][$key];
$filename = $_FILES['attach']['name'][$key];
$mail->AddAttachment($source, $filename);

}

将这些文件插入数据库位置的查询是

$insert = "INSERT INTO upload (name, type, size, file1, file2, file3, file4, file5,
    created) VALUES  
            ( 
            '".$name."',  
            '".$extention."',  
            '".$filesize."',  
            '".$MyUploads[0]."',  
            '".$MyUploads[1]."',  
            '".$MyUploads[2]."',  
            '".$MyUploads[3]."',
    '".$MyUploads[4]."' ,
    now() 
            )"; 
      $result = $dbLink->query($insert);

---

现在请告诉我如何在另一个php页面上检索这些文件。请告诉我，我是一个初学者，所以任何托图里亚语都是如此。

这已经在这里被回答了多次。。请看

编辑1

我觉得你的桌子应该是什么样子

CREATE TABLE IF NOT EXISTS `upload` (
  `uploadID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `uploadExt` varchar(10) NOT NULL,
  `uploadSize` varchar(50) NOT NULL,
  `uploadPath` varchar(200) NOT NULL,
  `uploadName` varchar(100) NOT NULL,
  PRIMARY KEY (`uploadID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

如何插入

$mysql = new mysqli("localhost","root","","test");
$sql = "INSERT INTO `upload` (
`uploadID` ,
`uploadExt` ,
`uploadSize` ,
`uploadPath` ,
`uploadName`
)
VALUES (
NULL ,  '%s',  '%s',  '%s',  '%s');"


foreach($image_you_have_uploaded as $valid) /// Am sure you know what this is 
{
    //Make sure you mysql_real_escape_string everthing 
    $mysql->query(sprintf($sql,$extention,$filesize,$filePath,$name));
} 

如何获取图像

$sql = "SELECT * FROM upload WHERE uploadName = '%s'" ;
$result = $mysql->query(sprintf($sql,mysql_real_escape_string($name)));

var_dump($result->fetch_assoc())// Output Result

---

谢谢。

谢谢你，在做了你所说的更改之后，我已经编辑了我的问题。请您详细说明我现在可以做些什么来检索它们我不同意您如何插入图像。。我会更新我认为最好的解决方案。。暂停相关：