How to upload multiple images at once using PHP

multiple images

[danjos]

I have a problem that i need to be helped with. Am trying to upload multiple images using one form and storing the images and thumbnails in their respective directories with the image names being stored in mysql database as image1, image2, image3 e.t.c but in the one table entry. My problem is that when i try to store the image names in the database the form inputs three table entries instead of one. My code is as follows :

[codes] <?
$max_no_img=3; // Maximum number of images value to be set here
echo "<form method=post action=car.php enctype='multipart/form-data'>";
echo "<table border='0' width='400' cellspacing='0' cellpadding='0' align=center>";
for($i=1; $i<=$max_no_img; $i++){
echo "<tr><td>Images $i</td><td>
<input type=file name='userfile[]' class='bginput'></td></tr>";
}
echo "<tr><td colspan=2 align=center><input type=submit value='Add Image'></td></tr>";

echo "</form> </table>";

?>[codes]

car.php
[codes]
<?php

$username = "";
$password = "";
$database = "";
$hostname = "";

mysql_connect($hostname,$username,$password);
mysql_select_db($database) or die( "Unable to select database");

function findexts ($filename)
{
$filename = strtolower($filename) ;
$exts = split("[/\\.]", $filename) ;
$n = count($exts)-1;
$exts = $exts[$n];
return $exts;
}



for($i=0; $i < 3; $i++)
{

$ext = findexts ($_FILES['userfile']['name'][$i]) ;
$ran = rand () ;
$ran2 = $ran.".";
$add = "upload/";
$add = $add . $ran2.$ext;
$image[$i] = $ran2.$ext ;
move_uploaded_file($_FILES['userfile']['tmp_name'][$i], $add);
$new_id = $image[$i]++;
$pic =$new_id++;

$query = "INSERT INTO table (image_name1,image_name2,image_name3)". "VALUES ('$ran.$ext','$new_id','$pic')";

mysql_query($query) or die('Database Query Error!');
}
echo "Successfully uploaded the image";



exit;

?>[codes]

[rocknbil]

A bit confusing, you say you want to store the image names multiple times but are setting different values for each field.

Anyway the problem is that your insert is inside the for loop, use the loop only to compile the query string. You sort out what you want to do with the three fields.

for($i=0; $i < 3; $i++)
{
mysql_query($query) or die('Database Query Error!');
}

Which inserts it three times. Make your loop just for compiling the recursive part of the select for the images, insert once.

Do something like this, compile the query into a single select. I'm only using the $ran.$ext value because I'm not fully getting why you're adding the other data when you say you want to insert the images in just the one field.

$query = "INSERT INTO table (image_name1,image_name2,image_name3) values(";

for($i=0; $i < 3; $i++) {
$ext = findexts ($_FILES['userfile']['name'][$i]) ;
$ran = rand () ;
$ran2 = $ran.".";
$add = "upload/";
$add = $add . $ran2.$ext;
$image[$i] = $ran2.$ext ;
move_uploaded_file($_FILES['userfile']['tmp_name'][$i], $add);
$new_id = $image[$i]++;
$pic =$new_id++;
$query .= "'$ran.$ext',";
}

//Strip off last comma
$query = preg_replace('/,$/','',$query);

$query .= ')';
mysql_query($query) or die('Database Query Error!');

[rocknbil]

Oops, popped back in, by query string I meant the select string stored in the variable $query, in case it's not obvious.