Multiple upload images PHP & Mysql

Question

Below code uploads images one by one. I want to upload multiple images!

    <?php
//config
        require("./configuration.php");

            $id = $_GET['id'];
            $q = mysql_query("SELECT * FROM books WHERE id='$id'") or die(mysql_error());
            $row = mysql_fetch_array($q);
            $name = $row['name'];
            echo '<div class="center-top"> NAME: <strong style="font-size:22px;">'.$name.'</strong>  </div>';
        //begin upload
        if(isset($_POST['submit'])) 
        {

            $title = htmlspecialchars($_POST['title']);
                $path = "./upload/images/";
                $name_pic = $_FILES['file']['name'];
                $ext = strtolower(substr(strrchr($name_pic, "."), 1));
                $allow = array("jpg", "jpeg", "JPG", "JPEG", "png", "gif");
               $uptype = ($_FILES['file']['tmp_name']);
                if (in_array($ext, $allow))
                {
                       $rand = rand(0,10000);
                       $md5  = md5($rand);
                       $new_file_name = "{$md5}.{$ext}";
                       $move_file = move_uploaded_file($_FILES['file']['tmp_name'], $path.$new_file_name);
                       if($move_file) {
                            mysql_query("INSERT INTO images (id, link, title, bid)
                            VALUES (NULL, '$new_file_name', '$title', '$id')") or die (mysql_error());
                            echo "<div class='yes'>succesfully added!</div>";   
        }}}

        //form
        echo '
        <form method="post" action="images.php?id='.$id.'"  enctype="multipart/form-data"> 
        <br />title: <br /><input type="text" name="title" value="'.$name.'"/><br />
        <br /><br/><input name="file" type="file" />
        <br /><input type="submit" name="submit" value="UPLOAD" /><br /><br />
        </form> 
        ';
        ?>

With this code delete an image:

<?php
$id = $_GET['id']; 
$result = mysql_query("DELETE FROM imagesWHERE `id`='$id'");
echo "<div class='yes'>successfully deleted picture!</div>";
?>

You are vulnerable to sql injection attacks. Enjoy having your server pwn3d. — Marc B
– Marc B, Commented Jan 25, 2016 at 16:37
@MarcB: Why do you have to be so rude about it? Give some positive feedback instead of this... — Chris G
– Chris G, Commented Jan 25, 2016 at 16:38
I myself upload images I'm admin, no other can access! @MarcB can you tell me what to do to jump over sql injection attack? — yaman
– yaman, Commented Jan 25, 2016 at 16:39
why does no one ever follow that link? it explains the problem and offers solutions for all kinds of languages... — Marc B
– Marc B, Commented Jan 25, 2016 at 16:43

Nana Partykar · Accepted Answer · 2016-01-25 17:40:59Z

0

First of all, you need to declare <input name="file" type="file"/> as <input name="file[]" type="file" multiple/> Then, find total nos of images selected. Put it in for loop and start storing in your upload->images folder and save image path in images column.

<?php

//config
require("./configuration.php");

$id = $_GET['id'];
$q = mysql_query("SELECT * FROM books WHERE id='$id'") or die(mysql_error());
$row = mysql_fetch_array($q);
$name = $row['name'];
echo '<div class="center-top"> NAME: <strong style="font-size:22px;">'.$name.'</strong>  </div>';
if(isset($_POST['submit'])) 
{
    $title = htmlspecialchars($_POST['title']);
    $path = "./upload/images/";
    $TotalImage = count($_FILES['file']['name']);
    for($i=0;$i>$TotalImage;$i++)
    { 

        $image_name = $_FILES['file']['name'][$i];
        $new_file_name = $path.$image_name;
        if(move_uploaded_file($_FILES['file']['tmp_name'][$i],$path.$image_name))
        {
            mysql_query("INSERT INTO images (id, link, title, bid) VALUES (NULL, '$new_file_name', '$title', '$id')") or die (mysql_error());
            echo "<div class='yes'>succesfully added!</div>";   
        }
    }
}

//form
echo '
<form method="post" action="images.php?id='.$id.'"  enctype="multipart/form-data"> 
<br />title: <br /><input type="text" name="title" value="'.$name.'"/><br />
<br /><br/><input name="file[]" type="file" multiple/>
<br /><input type="submit" name="submit" value="UPLOAD" /><br /><br />
</form> 
';
?>

answered Jan 25, 2016 at 17:40

Nana Partykar

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2 Comments

yaman Over a year ago

Actually doesn't work! No errors but not inserts into database

Nana Partykar Over a year ago

You have used the same code which i wrote as an answer. Because, in my desktop it's working. Lot of differences is there in your or my code. So, check it properly. It should work. @yaman. Not only input file changed but, also the execution page also changed. Check it once.

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