Question

我问如何将多个文件（使用jquery.Multifile）上传到FTP并将文件链接到MySQL数据库？我已经有了一个HTML表单，可以使用PHP将多个文件上传到我的FTP。

假设我有两个表（我在这些表中也有其他字段但是将它们添加到我的问题中无关紧要）：

讯息 - ＆gt; messageID（pk），标题，日期，消息

文件 - ＆gt; fileID（pk），messageID（fk），fileLink

当用户编写消息时，他可以选择许多文件作为jquery.Multifile的附件。然后，当用户发送消息时，文件被发送到FTP，并且文件的http链接被发送到数据库。每个文件都应该有自己的ID（fileID），这两个表使用字段messageID作为消息表中的主键和文件表中的外键链接在一起。

这就是我现在所拥有的，但我不知道如何处理数据库部分：

HTML表单：

<form name="uploader" action="file_upload2.php" method="post" enctype="multipart/form-data">
<fieldset>
    <label for="file">Upload file:</label>
    <br />
        <input type="file" class="multi, remove" name="file[]" value="Upload file" />
        <input type="submit" value="submit"/>
</fieldset>

这是PHP部分（$ filepath是我试图在FTP中建立文件的链接）：

<?php
include('db.php');

mysql_connect("$hostname", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$database")or die("cannot select DB");
mysql_query("SET names 'UTF8'");

$heading=$_POST['heading'];
$date=$_POST['date'];
$message=$_POST['message'];

$conn = ftp_connect("mysite.org") or die("Cannot connect");
ftp_login($conn,"username","password");
$uploaddir = 'httpdocs/folder1/folder2/files/';

foreach ($_FILES["file"]["error"] as $key => $error)
{
        if ($error == UPLOAD_ERR_OK)
        {
                $tmp_name = $_FILES["file"]["tmp_name"][$key];
                $name = $_FILES["file"]["name"][$key];
                $uploadfile = $uploaddir . basename($name);
                ftp_put($conn,$uploadfile,$tmp_name,FTP_BINARY);

            if (move_uploaded_file($tmp_name, $name))
            {
                    echo "Success: File ".$name." uploaded to server.<br/>";
            }
            else
            {
                    echo "Error: File ".$name." can not be uploaded to server.<br/>";
            }
    }
}

ftp_close($conn);

$filepath = "http://mysite.org/folder1/folder2/files/".$name;

$sql="INSERT INTO message(
heading, 
date, 
message)
VALUES
(
 '$heading',
 '$date',
 '$message')";

$result=mysql_query($sql);

if($result){
header("Location: ../index.php");

}
else {
    echo $sql . ("Adding message failed <br />" . mysql_error());
}
mysql_close();
?>

Answer 1

试试这个

<?php

include('db.php');

mysql_connect("$hostname", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$database")or die("cannot select DB");
mysql_query("SET names 'UTF8'");

$heading=$_POST['heading'];
$date=$_POST['date'];
$message=$_POST['message'];

$conn = ftp_connect("mysite.org") or die("Cannot connect");
ftp_login($conn,"username","password");
$uploaddir = 'httpdocs/folder1/folder2/files/';

$sql="INSERT INTO message(messageid,heading, date, message) VALUES('Null' , '$heading', '$date', '$message')";
$result=mysql_query($sql);
$lastinsertid = mysql_insert_id($sql);

foreach ($_FILES["file"]["error"] as $key => $error)
{
        if ($error == UPLOAD_ERR_OK)
        {
                $tmp_name = $_FILES["file"]["tmp_name"][$key];
                $name = $_FILES["file"]["name"][$key];
                $uploadfile = $uploaddir . basename($name);
                ftp_put($conn,$uploadfile,$tmp_name,FTP_BINARY);

            if (move_uploaded_file($tmp_name, $name))
            {
                    echo "Success: File ".$name." uploaded to server.<br/>";
                    $filepath = "http://mysite.org/folder1/folder2/files/".$name;
                    $sql="INSERT INTO File (fileid, messageid, link) VALUES( Null, '$lastinsertid', '$filepath')";
                    $result=mysql_query($sql);


            }
            else
            {
                    echo "Error: File ".$name." can not be uploaded to server.<br/>";
            }
    }
}

ftp_close($conn);




if($result){
header("Location: ../index.php");

}
else {
    echo $sql . ("Adding message failed <br />" . mysql_error());
}
mysql_close();
?>

将多个文件上传到FTP和HTTP链接到MySQL数据库的文件

1 个答案: