2012-12-25 49 views
3

我有一個能夠將多個選定文件上傳到該文件夾​​的表單。我只需要找到如何使用PHP將它們插入到數據庫中?如何將許多圖像添加到MySQL數據庫?

HTML表單

<form name="demoFiler" id="demoFiler" enctype="multipart/form-data"> 
<input type="file" name="multiUpload" id="multiUpload" multiple /> 
<input type="submit" name="submitHandler" id="submitHandler" value="Upload" class="buttonUpload" /> 
</form> 

PHP代碼

if($_SERVER['REQUEST_METHOD'] == "POST"){ 
    if(move_uploaded_file($_FILES['file']['tmp_name'], "uploads/".$_FILES['file']['name'])){ 
     echo($_POST['index']); 
    } 
    exit; 
} 

這是我的代碼,並首先我不能得到的文件被移動到文件夾上傳:

<?php 

$username = "root"; 
$password = ""; 
$host = "localhost"; 
$database = "db_holiday"; 

// Make the connect to MySQL or die 
// and display an error. 
$link = mysql_connect($host, $username, $password); 
if (!$link) { 
    die('Could not connect: ' . mysql_error()); 
} 

// Select your database 
mysql_select_db ($database); 

if($_SERVER['REQUEST_METHOD']=="POST") 
{ 
    foreach ($_FILES['multiUpload']['name'] as $fileName) { 
     $uploaddir= '/upload/'; 
    $uploadfile = $uploaddir . basename($fileName); 
     if (move_uploaded_file($fileName, $uploadfile)) { 
      echo "File is valid, and was successfully uploaded.\n"; 
     } else { 
      echo "Possible file upload attack!\n"; 
     } 
    } 
} 

?> 
<form action="" method="POST" enctype="multipart/form-data"> 
<input type="file" name="multiUpload[]" id="multiUpload" multiple="multiple" /> 
<input type="submit" name="button"/> 
</form> 
+3

這取決於你想如何保存圖像?你想將它保存爲glob,服務器上某處文件的路徑,base64編碼的圖像嗎?我會親自爲每次上傳創建一個獨特的文件夾,並將該路徑與數據庫中的圖像相關聯, – thescientist

回答

0

使用2 input type="file"元素

<input type="file" name="img_1" /> 
<input type="file" name="img_2" /> 

訪問的每個文件使用

$_FILES['img_1'][]; 
$_FILES['img_2'][]; 

或者你也可以像使用

<input type="file" name="img[]" /> 
<input type="file" name="img[]" /> 
<!--Which is often used in checkboxes--> 

數組和訪問使用

$_FILES['img']['tmp_name'][0]; 
$_FILES['img']['tmp_name'][1]; 
+0

原因-1請 –

+0

-1。這個答案中沒有提到循環,在討論**許多**圖像時,這是真正的要求。 –

+0

@TomvanderWoerdt不明白?我的意思是我們現在需要提供飼料嗎? –

4

HTML每個圖像:

<input type="file" name="multiUpload[]" id="multiUpload" multiple /> 

PHP:

foreach ($_FILES['multiUpload'] as $fileName) { 
    //write your code here 
    //$fileName 
}