简体   繁体   English

如何将多个上传文件保存到数据库

[英]How to save multiple upload file to the database

I have a problem to save multiple upload file to the database. 将多个上传文件保存到数据库时出现问题。 i use drag and drop multiple upload file with html 5. for example, i want to save 2 image in the 1 table but different row. 我将拖放的多个上传文件与html 5配合使用。例如,我想在1个表中保存2张图像,但另存一行。 so, this is my simple structure table database and web. 因此,这是我的简单结构表数据库和Web。

table database 表数据库

name, 'int', '25'.
image1, 'var', '100'.
image2, 'var', '100'.

index.php index.php

<!DOCTYPE html>
<html>
<head>
  <title>Page Title</title>


<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript" src="multiupload.js"></script>
<script type="text/javascript">
var config = {
// Valid file formats
support : "image/jpg,image/png,image/bmp,image/jpeg,image/gif",
form: "demoFiler", // Form ID
dragArea: "dragAndDropFiles", // Upload Area ID
uploadUrl: "upload.php" // Server side file url
}
//Initiate file uploader.
$(document).ready(function()
{
initMultiUploader(config);
});
</script>

</head>

<body>

 <div id="dragAndDropFiles" class="uploadArea">
<h1>Drop Images Here</h1>
</div>
<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" />
</form>
<div class="progressBar">
<div class="status"></div>
</div>

</body>

</html> 

upload.php upload.php

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

So, what i must to do for modify upload.php script?? 所以,我必须做什么来修改upload.php脚本? thank's for you're advice.. 感谢您的建议。

You can use your form like this : 您可以这样使用表单:

First you need to change your form select into: name="multiUpload[]" 首先,您需要将表单选择更改为: name="multiUpload[]"

$arr = $_GET['multiUpload'];

foreach($arr as $option) {
    echo $option."<br>"; //Do whatever you want.
}

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM