Uploading MANY Pictures!

**JafferAB** · Jul 5th, 2002, 11:22 AM

Im in the process of writing a script which will upload many files. So far i have created a bit that will give me how many forms i need to upload however many files i need, but it doesnt upload them, or move them to another directory...

I have looked through the forums and all the upload thingds only relate to doing one file...

here my scripty!

PHP Code:


if($page=="4") {

    ?>

    <html><head><title>New Picture Section - Part 2</title></head>

    <body bgcolor="#CCCCCC">

    <font face="verdana" size="5">Add new Pictures to <?=$secname?><font size="2"><BR><BR>Upload Pictures and Info. This page will problalbly take you a long time, so i would get something downloading while your not using the net (as such :D). There is no reset button on here because if you hit it accidentally, you would be totally devistated :D:D<BR><BR>



    <?PHP



    if (isset($_POST['submit'])) {



    while($y>1) {



        print "this code is running<BR>";



        $noofpic = $number;

        $big = "big";

        $small = "small";



      $fileUploadCopy = $_FILES['$y$small']['tmp_name'];

      $fileUpload = $_FILES['$y$small']['name'];

      $fileUploadCopy2 = $_FILES['$y$big']['tmp_name'];

      $fileUpload2 = $_FILES['$y$big']['name'];



      $imageName = $fileUpload;

      $imageName2 = $fileUpload2;



        $y--;



      if ($fileUploadCopy != "none") {

        copy($fileUploadCopy,"uploads/$imageName");

    }

      if ($fileUploadCopy2 != "none") {

        copy($fileUploadCopy2,"uploads/$imageName");

    }

      }

      }



    if(!$thumbnails==NULL && !$textforeach==NULL) {



        $noofpic = $number;

        $y = 0;

        $big = "big";

        $small = "small";



        #print "<table>";

        print "<form method=\"post\" enctype=\"multipart/form-data\">";

        #print "<input type=\"hidden\" name=\"page\" value=\"5\">";



        while($y < $noofpic) {

            $s = $y + 1;

            print "<table><TR><TD>Picture no.</td><td>$s</td></tr><tr><td>Thumbnail</td><td><input type=\"file\" name=\"$y\"></td></tr><tr><td>Big Picture</td><td><input type=\"file\" name=\"$y$small\"></td></tr><tr><td>Text</td><td><input type=\"text\" name=\"$y$big\"></td></tr></table><BR>";

            $y++;

        }



        print "<input type=\"submit\" name=\"submit\" value=\"Upload\">";

        print "</form>";



    }



else print "There has been an error while executing this script. Please retry, or send an email to the <a href=\"mailto:$webmasteremail?subject=Error with picture page\">webmaster</a>";

}

?>

as you can see i have used while loops and stuff, and tryed to double the amount of stuff you can upload, any help is appreciated, thanks

**cpradio** · Jul 5th, 2002, 12:45 PM

Make sure the directory you are uploading to is chmoded to 777

**JafferAB** · Jul 5th, 2002, 12:52 PM

this is my own server so i have full access to everyhting :/

**cpradio** · Jul 5th, 2002, 01:13 PM

Are you on a NT or Unix server?

**JafferAB** · Jul 5th, 2002, 01:43 PM

NT, win2k, running Apache

It does work because i tried a test someone else posted on the board (it was you)

PHP Code:


<?PHP



if (isset($_POST['submit'])) {

  $fileUploadCopy = $_FILES['filename']['tmp_name'];

  $fileUpload = $_FILES['filename']['name'];

  $imageName = $fileUpload;

  if ($fileUploadCopy != "none")

    copy($fileUploadCopy,"uploads/$imageName");

} else {

  echo "<form method=\"post\" action=\"?\" enctype=\"multipart/form-data\">" .

          "<input type=\"file\" name=\"filename\">" .

          "<input type=\"submit\" value=\"upload\" name=\"submit\">";

}



?>

That works and was the basis for my script

**cpradio** · Jul 5th, 2002, 01:52 PM

take off the quotes around the variables:

PHP Code:


      $fileUploadCopy = $_FILES['$y$small']['tmp_name'];

      $fileUpload = $_FILES['$y$small']['name'];

      $fileUploadCopy2 = $_FILES['$y$big']['tmp_name'];

      $fileUpload2 = $_FILES['$y$big']['name'];

should be:

PHP Code:


      $fileUploadCopy = $_FILES[$y$small]['tmp_name'];

      $fileUpload = $_FILES[$y$small]['name'];

      $fileUploadCopy2 = $_FILES[$y$big]['tmp_name'];

      $fileUpload2 = $_FILES[$y$big]['name'];

-Matt

**JafferAB** · Jul 5th, 2002, 02:00 PM

hmmm, it doesnt like that atall

Parse error: parse error, unexpected T_VARIABLE, expecting ']' in C:\Program Files\Apache Group\Apache2\htdocs\PHP\MySite\admin\admin.php on line 149

when they were inside ' ' s it gives no error

**JafferAB** · Jul 6th, 2002, 04:17 AM

is it important that the fileupload vvariable is called fileupload? i mean it can be called fileupload 2 cant it...

i wonder why i cant take the $y$big out of the ''s.

**cpradio** · Jul 6th, 2002, 08:20 AM

I see your problem now. $y$big is not the name of your second file. Your two file names are $y and $y$small

**JafferAB** · Jul 6th, 2002, 10:18 AM

ahhhhhh, its not only that

while tryign to create another script of the same type i ran into the problem

The problem is that i am trying to use variables cleverly, my new script...

PHP Code:


if($section=="0") {

        $y = 0;

        $text = "text";

        $target= "target";



        print "<form name=\"linkys\" action=\"admin.php\">";

        print "<input type=\"hidden\" name=\"page\" value=\"7\">";

        print "<input type=\"hidden\" name=\"section\" value=\"2\">";

        print "<input type=\"hidden\" name=\"numberof\" value=\"$numberof\">";





        while($y < $numberof) {

            $s = $y + 1;

            print "<table><TR><TD>Link no.</td><td>$s</td></tr><tr><td>URL</td><td><input type=\"text\" name=\"$y\"></td></tr><tr><td>URL Text</td><td><input type=\"text\" name=\"$y$text\"></td></tr><tr><td>New Window</td><td><input type=\"checkbox\" name=\"$y$target\"></td></tr></table><BR>";

            $y++;

        }



        print "<input type=\"submit\" value=\"Change Links\">";

        print "</form>";

    }



    if($section=="2") {



        $y = 0;

        $t = $numberof;

        $target = "target";

        $text = "text";



        $filename = "links.txt";

        $fp = fopen($filename,'w');

        fwrite($fp,"");

        fclose($fp);



        while($y < $t) {



            $ytext = $y.text;



            $filename = "links.txt";

            $fp = fopen($filename,'a');

            fwrite($fp,"<a href=\"$1\"");



            if ($y.$target != NULL) {

                fwrite($fp," target=\"_NEW\">$$ytext</a>");

            }

            else {

                fwrite($fp,">$y.$text</a>");

            }



            fclose($fp);



            $y++;



        }



    }

Trys to dynamically create variables too, but when it comes to writing to a file, all i get it...

<a href="$1" target="_NEW">$0text</a><a href="$1" target="_NEW">$1text</a>

so all i nee dnow is a way around this, so that synamically created variablesa will work when tryed to run under a new condition (it ihnk)

**nabeels786** · Jul 6th, 2002, 02:16 PM

with something like:

"<a href=\"$1\""

try

"<a href=\"" . $1 . "\""

i think this is what u need

PHP Code:


$fp = fopen($filename,'a');

            fwrite($fp,"<a href=\"" . $1 . "\"");



            if ($y.$target != NULL) {

                fwrite($fp," target=\"_NEW\">" . $y . $text . "</a>");

            }

            else {

                fwrite($fp,">" . $y . $text . "</a>");

            }

**nabeels786** · Jul 6th, 2002, 05:06 PM

then its the values of the variables.

you have

try changing:
$y++;

to

$y+=1;

change

if($section=="2") {

to

elseif($section=="2") {

what is $ytext = $y.text; ?

**nabeels786** · Jul 6th, 2002, 05:15 PM

try this

PHP Code:


$y = 0;

$text = "text";

$target= "target";

$t = $numberof;



if($section=="0") {

       print "<form name=\"linkys\" action=\"admin.php\">";

    print "<input type=\"hidden\" name=\"page\" value=\"7\">";

    print "<input type=\"hidden\" name=\"section\" value=\"2\">";

    print "<input type=\"hidden\" name=\"numberof\" value=\"$numberof\">";





    while($y < $numberof) {

        $s = $y + 1;

        print "<table><TR><TD>Link no.</td><td>" . $s . "</td></tr><tr><td>URL</td><td><input type=\"text\" name=\"" . $y . "\"></td></tr><tr><td>URL Text</td><td><input type=\"text\" name=\"" . $y . $text . "\"></td></tr><tr><td>New Window</td><td><input type=\"checkbox\" name=\"" . $y . $target . "\"></td></tr></table><BR>";

        $y+=1;

    }



    print "<input type=\"submit\" value=\"Change Links\">";

    print "</form>";

} elseif($section=="2") {

    $filename = "links.txt";

    $fp = fopen($filename,'w');

    fwrite($fp,"");

    fclose($fp);



    while($y < $t) {



        $ytext = $text;



        $filename = "links.txt";

        $fp = fopen($filename,'a');



        if ($y.$target != NULL) {

            fwrite($fp,"<a href=\"" . $1 . "\"  target=\"_NEW\">" . $y . $text . "</a>");

        }  else {

            fwrite($fp,"<a href=\"" . $1 . "\">" . $y . $text . "</a>");

        }



        fclose($fp);



        $y+=1;



    }



}

is the form on top submitting to itself?
because the $text var is the same as the name of one of the form fields.

**cpradio** · Jul 6th, 2002, 05:18 PM

This is just my opinion but if I were writing this I would make the form so it used an array and not just adding a 0text to the name.

Something like this:

PHP Code:


print "<table><tr><td>Link no.</td><td>" . $s . "</td></tr>" .

       "<tr><td>URL</td><td><input type=\"text\" name=\"url[]\"></td></tr>" .

       "<tr><td>URL Text</td><td><input type=\"text\" name=\"urlText[]\"></td></tr>" .

       "<tr><td>New Window</td><td><input type=\"checkbox\" name=\"window[]\"></td></tr></table><BR>";

**JafferAB** · Jul 6th, 2002, 06:11 PM

AN ARRAY, YOUR A GENIOUS!!!!!!

/me kissed cpradio

**JafferAB** · Jul 6th, 2002, 06:28 PM

ok, i got this...

PHP Code:


if($page=="4") {

    ?>

    <html><head><title>New Picture Section - Part 2</title></head>

    <body bgcolor="#CCCCCC">

    <font face="verdana" size="5">Add new Pictures to <?=$secname?><font size="2"><BR><BR>Upload Pictures and Info. This page will problalbly take you a long time, so i would get something downloading while your not using the net (as such :D). There is no reset button on here because if you hit it accidentally, you would be totally devistated :D:D<BR><BR>



    <?PHP



    if (isset($_POST['submit'])) {



    while($y>1) {



        print "this code is running<BR>";



        $noofpic = $number;



        $fileUploadCopy = $_FILES['thumbnail[$y]']['tmp_name'];

          $fileUpload = $_FILES['thumbnail[$y]']['name'];

          $fileUploadCopy2 = $_FILES['bigpics[$y]']['tmp_name'];

          $fileUpload2 = $_FILES['bigpics[$y]']['name'];



      $imageName = $fileUpload;

      $imageName2 = $fileUpload2;



        $y--;



      if ($fileUploadCopy != "none") {

        copy($fileUploadCopy,"uploads/$imageName");

    }

      if ($fileUploadCopy2 != "none") {

        copy($fileUploadCopy2,"uploads/$imageName");

    }

      }

}



    if(!$thumbnails==NULL && !$textforeach==NULL) {



        $noofpic = $number;

        $y = 0;

        #$thumbnails[];

        #$smallpics[];

        #$text[];



        print "<form method=\"post\" enctype=\"multipart/form-data\">";



        while($y < $noofpic) {

            $s = $y + 1;

            print "<table>

            <TR><TD>Picture no.</td><td>$s</td></tr>

            <tr><td>Thumbnail</td><td><input type=\"file\" name=\"thumbnail[$y]\"></td></tr>

            <tr><td>Big Picture</td><td><input type=\"file\" name=\"bigpics[$y]\"></td></tr>

            <tr><td>Text</td><td><input type=\"text\" name=\"text[$y]\"></td></tr></table><BR>";

            $y++;

        }



        print "<input type=\"submit\" name=\"submit\" value=\"Upload\">";

        print "</form>";



    }



else print "There has been an error while executing this script. Please retry, or send an email to the <a href=\"mailto:$webmasteremail?subject=Error with picture page\">webmaster</a>";

}

I tihnk i can problalbly work it out from here, as long as the variables work, which si the only thing im having trouble working out if they actually are doing :-/

**JafferAB** · Jul 6th, 2002, 06:54 PM

you know what, ive completele lost what the hell im talking about :/

this is the prblem with PHP, once you get some variable sinto play, a few loops and a big script, WHERE THE HELL IS WHAT YOUR LOOKING FOR

**The Hobo** · Jul 7th, 2002, 07:35 AM

Originally posted by nabeels786
with something like:

"<a href=\"$1\""

try

"<a href=\"" . $1 . "\""

i think this is what u need

They are the same thing.

Originally posted by nabeels786

$y++;

to

$y+=1;

And so are they. The way he is doing it is the more efficeint (correct) way of doing it. You're telling him to go from good programming to elementary.

**JafferAB** · Jul 7th, 2002, 08:18 AM

yeah. And all im really trying to do first is get it working

but as i say, all the script is confusing me now, i outta isolate it really and think about ti a bit

**JafferAB** · Jul 7th, 2002, 08:38 AM

HAHA, IM A GENIOUS (allthough technically the idea was cpraio's

)

PHP Code:


if($section=="0") {



        $y = 0;



        print "<form name=\"linkys\" action=\"admin.php\">";

        print "<input type=\"hidden\" name=\"page\" value=\"7\">";

        print "<input type=\"hidden\" name=\"section\" value=\"2\">";

        print "<input type=\"hidden\" name=\"numberof\" value=\"$numberof\">";





        while($y < $numberof) {



            $s = $y + 1;



            print "<table><TR><TD>Link no\.</td><td>$s</td></tr>";

            print "<tr><td>URL</td><td><input type=\"text\" name=\"url[$y]\"></td></tr>";

            print "<tr><td>URL Text</td><td><input type=\"text\" name=\"urltext[$y]\"></td></tr>";

            print "<tr><td>New Window</td><td><input type=\"checkbox\" name=\"newwindow[$y]\"></td></tr>";

            print "</table><BR>";



            $y++;



        }



        print "<input type=\"submit\" value=\"Change Links\">";

        print "</form>";

    }



    if($section=="2") {



        $y = 0;

        $t = $numberof;



        $filename = "../texts/links.txt";

        $fp = fopen($filename,'w');

        fwrite($fp,"");

        fclose($fp);



            while($y < $t) {



                $filename = "../texts/links.txt";

                $fp = fopen($filename,'a');

                        fwrite($fp,"<a href=\"".$url[$y]."\"");



                        if ($newwindow[$y] != NULL) {

                            fwrite($fp," target=\"_NEW\">".$urltext[$y]."</a>");

                            }

                        else {

                            fwrite($fp,">".$urltext[$y]."</a>");

                        }



                fclose($fp);



                $y++;



            }

righty ho, my links creator now WORKS WITH dynamic link adding abilities

time to do the same for the file upload script

**JafferAB** · Jul 7th, 2002, 09:12 AM

im OH SO CLOSE to the pictures script running, just one thing i havent ocme accross, and neither has me book

PHP Code:


if ($page=="5") {



    if (isset($_POST['submit'])) {



    $y = 0;

    $s = (count($thumbnail)-1);



        while($s>$y) {



            print "this code is running<BR>";

            $noofpic = $number;



            $fileUploadCopy = $_FILES['thumbnail[$y]']['tmp_name'];

             $fileUpload = $_FILES['thumbnail[$y]']['name'];

             $fileUploadCopy2 = $_FILES['bigpics[$y]']['tmp_name'];

              $fileUpload2 = $_FILES['bigpics[$y]']['name'];



            $imageName = $fileUpload;

            $imageName2 = $fileUpload2;



            $y++;



            if ($fileUploadCopy != "none") {

              copy($fileUploadCopy,"uploads/$imageName");

            }

              if ($fileUploadCopy2 != "none") {

                 copy($fileUploadCopy2,"uploads/$imageName");

            }

        }

        print "woo";

    }

}

creates this error....

Warning: Unable to open '' for reading: Permission denied in C:\Program Files\Apache Group\Apache2\htdocs\PHP\MySite\admin\admin.php on line 203

Warning: Unable to open '' for reading: Permission denied in C:\Program Files\Apache Group\Apache2\htdocs\PHP\MySite\admin\admin.php on line 206

i know it means it has no permission to open and upload the file, or copy it or whatever, but does anyone know how to get permission?

Thanks you

**nabeels786** · Jul 7th, 2002, 02:27 PM

Originally posted by The Hobo

They are the same thing.

And so are they. The way he is doing it is the more efficeint (correct) way of doing it. You're telling him to go from good programming to elementary.

i know its the same, i was thinking maybe it was screwing up on that.

**nabeels786** · Jul 7th, 2002, 02:28 PM

chmod on the folders/

**JafferAB** · Jul 7th, 2002, 05:06 PM

well, i dont think CHMODing will help, its my server and my files, and besides i ddidnt think you needed to CHMOD to read anyway, it shouldnt be that, but i will try anyway..

Another thing is i have a problem, my monitor set on fire today so im using my TV whihc as some of you may know is extremele low quality for text

i can watch some nice films hopwever, anyway i might have to stop work on this, ill keep plugging till the pic script is done though

**The Hobo** · Jul 7th, 2002, 05:25 PM

Originally posted by nabeels786
chmod on the folders/

You can't chmod on NT servers. Only on UNIX.

**JafferAB** · Jul 7th, 2002, 05:56 PM

indeed

**cpradio** · Jul 7th, 2002, 06:28 PM

you need to set the folder permissions. So it can read, execute, and write.

**The Hobo** · Jul 7th, 2002, 06:31 PM

Originally posted by cpradio
you need to set the folder permissions. So it can read, execute, and write.

Maybe I'm confused, but didn't he already say that he did?

**cpradio** · Jul 7th, 2002, 06:55 PM

oops sorry, I really did not read too much of this thread recently, the problem is either you have mispelled a variable name or the form is not sending the information correctly because your copy() functions think the file you are trying to upload is '' (blank)

**JafferAB** · Jul 7th, 2002, 09:04 PM

cool, well at least this means im nearly there

no doubt i will finish it tommorrow

i tihnk then, that what your saying is all i have to do is get the program to read the variable correctly ? that shouldnt be too hard now im using arrays, i tryed using them on the links making page and it worked a treat

thanks fro all j00 help

**Kagey** · Jul 10th, 2002, 12:32 PM

try changing this:

PHP Code:


$fileUploadCopy = $_FILES['thumbnail[$y]']['tmp_name'];

             $fileUpload = $_FILES['thumbnail[$y]']['name'];

             $fileUploadCopy2 = $_FILES['bigpics[$y]']['tmp_name'];

              $fileUpload2 = $_FILES['bigpics[$y]']['name'];

to this (double quoted)

PHP Code:


$fileUploadCopy = $_FILES["thumbnail[$y]"]['tmp_name'];

             $fileUpload = $_FILES["thumbnail[$y]"]['name'];

             $fileUploadCopy2 = $_FILES["bigpics[$y]"]['tmp_name'];

              $fileUpload2 = $_FILES["bigpics[$y]"]['name'];

scoutt · Jul 10th, 2002, 04:52 PM

kagey that will error out. you can't use " in the globals.

Jaffer you are making thi shard on yourself. for one you can't use numbers as variables. $1 does not work.

also if I'm not mistaken the $_FILES array looks like this

$_FILES['thumbnail']['tmp_name'][$y];

let me check and also see if I can come up with some thing better.

also you migth want to try to add the whole path

/path/to/uploads/$imageName

**Kagey** · Jul 10th, 2002, 05:37 PM

scoutt's way of referencing the uploaded file array works. i am trying to do the same thing, and i just found an example in the manual.

here is a php bit that will work. assuming that the picture upload fields were called pictures[]:

PHP Code:


$nPics = (count($pictures));



for($i = 0; $i < $nPics; $i++) {



    if(stristr($HTTP_POST_FILES['pictures']['type'][$i], 'image')){ // only accept picture files

    

        if (is_uploaded_file($HTTP_POST_FILES['pictures']['tmp_name'][$i]))

            copy($HTTP_POST_FILES['pictures']['tmp_name'][$i], 

            "somepath/" . basename($HTTP_POST_FILES['pictures']['name'][$i]));

    }

}

scoutt · Jul 10th, 2002, 07:43 PM

if you set up the inputs in array like so

<input type="file" name="uploadedfile[]">
<input type="file" name="uploadedfile[]">
<input type="file" name="uploadedfile[]">

you could do something like so

PHP Code:


<?php

if (isSet($_REQUEST['submit']))

{

$file_counter = 0;

$i = 0;

while (list ($key, $file) = each ($_FILES))

{

if ($file['tmp_name'] != "none")

{

$uploaded_tmp_name = $file['tmp_name'];

$uploaded_name = $file['name'];



if (!copy($uploaded_tmp_name, $upload_dir.$uploaded_name))

{

echo " There was an ERROR when moving the file <b>\"$uploaded_name\"</b> to <b>\"$upload_dir\"!</b><br>";

}

else

{

echo " A file <b>\"$uploaded_name\"</b> was moved to <b>\"$upload_dir\"</b> !<br>";

}

?>

you also have to remember what version of php you are trying to work woth as if it is 4.1 or 4.2 I don't think is_uploaded_file() works anymore.

**JafferAB** · Jul 11th, 2002, 03:59 PM

IT WORKS

HALE, THE CONQUERING HERO!!!

DA DA DADADADA, DA DAAAAAAAAAAAAAAAAAAAAAAAAAA

i am the mata

when seeing your posts i thought that my code had to be right, i rewrote it loadas and loads! and then, THEM I REALISED, it wasnt passing $number because i kept refreshing, $number was potentially lost in the second stage, SO, i went back and put in hidden input field in the second stage, this notched it up and let the loop run,t hus uploading some files

HARDCORE

Thnak you all so much for helping me, i would have given up if people hadnt of kept tryign to give me new thinga to do

**JafferAB** · Jul 11th, 2002, 04:09 PM

oh, and if your pondering my script, its...

PHP Code:


if($page=="3") {

    ?>

    <html><head><title>New Picture Section - Part 1</title></head>

    <body bgcolor="#CCCCCC">

    <font face="verdana" size="5">Add new Pictures<font size="2"><BR><BR>Add new Section of Pictures<BR><BR>

    <?PHP



    if($section=="2") {

    ?>

    <form name="getinfo" action="admin.php">

    <input type="hidden" name="page" value="3">

    <input type="hidden" name="section" value="0">

    <TABLE>

    <TR><TD>Thumbnails</td><td><input type="checkbox" name="thumbnails" checked><font size="1">Please note that NOT using this will mean your BIG image will be compressed to a thumbnail size</TD></TR>

    <TR><TD>Text for each Pic</td><td><input type="checkbox" name="textforeach" checked></TD></TR>

    <tr><td>New Section Name</td><td><input type="text" name="secname"></td></tr>

    <tr><td>Section Welcome Text</td><td><textarea cols="40" rows="5" name="welcome text"></textarea></td></tr>

    <tr><td>Number of Pictures</td><td><input type="text" name="number"></td></tr>

    <tr><td>Background Color (Pic Page)</td><td><input type="text" name="piccolor" value="#"></td></tr>

    <tr><td></td><td><input type="submit" value="Next Section"></td></tr>

    </table></form>

    </body></html>



    <?PHP

    }

    if($section == "0") {



        $y=0;



        print "<form name=\"cool\" method=\"POST\" action=\"admin.php\" enctype=\"multipart/form-data\">";

        print "<input type=\"hidden\" name=\"page\" value=\"3\">";

        print "<input type=\"hidden\" name=\"section\" value=\"1\">";

        print "<input type=\"hidden\" name=\"number2\" value=\"$number\">";

        print "<table>";



        while ($y < $number) {

            print "<tr><td>Link No.</td><td>$y</td></tr>";

            print "<tr><td>Thumnail Link</td><td><input type=\"file\" name=\"thumbnail[$y]\"></td></tr>";

            print "<tr><td>BigPic Link</td><td><input type=\"file\" name=\"bigpic[$y]\"></td></tr>";

            print "<tr><td>Text</td><td><input type=\"text\" name=\"text[$y]\"></td></tr>";

            $y++;

        }



        print "<tr><td></td><td><input type=\"submit\" name=\"submit\" value=\"Upload!\">";

        print "</table></form>";

        }

        }



    if($section=="1") {

        if (isset($_POST['submit'])) {



            $y=0;

            $number = $number2;



            while ($y < $number) {

                  $fileUploadCopy = $_FILES['thumbnail']['tmp_name'][$y];

                  $fileUpload = $_FILES['thumbnail']['name'][$y];



                  print $fileupload;



                  $imageName = $fileUpload;

                  if ($fileUploadCopy != "none") {

                    copy($fileUploadCopy,"../pictures/uploaded/$imageName");

                }



                 $fileUploadCopy = $_FILES['bigpic']['tmp_name'][$y];

                 $fileUpload = $_FILES['bigpic']['name'][$y];



                 $imageName = $fileUpload;

                 if ($fileUploadCopy != "none") {

                   copy($fileUploadCopy,"../pictures/uploaded/$imageName");

                }



                $y++;

            }

        }

}

**cpradio** · Jul 11th, 2002, 05:13 PM

tsk, tsk, tsk. you broke most of the 4.2 standards

**JafferAB** · Jul 11th, 2002, 06:01 PM

bah!!! as long as it works im happy

**the_joker** · Jan 30th, 2003, 01:35 AM

ok, im a complete newbie with a little knowledge of php

but i dont get how the last script posted is supposed to work
how can you start with an if statement without the "<?" ?
im not sure what to change to be able to upload to the server im working with

**phpman** · Jan 30th, 2003, 10:42 AM

he just forgot it. he must have more code above that that has the open <?

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