how do you do that ??

hello, I have this line in my upload script code

        #If link is to be shown, show link

        if ($see_url)

        {

      	  $uploaderinfo .= $f_link . "<br><a href=\"" . $url_path . "/uploads27/" . $newfilename . "\" target=\"_blank\">" . $url_path . "/uploads27/" . $newfilename . "</a><br><br>\n";

        }

PHP:

This part of teh code shows the link of the file uploaded.

I would like to add below it the following forms with the links in it but they are in html ,, when i place them i get an error in php , how do i place the following code correctly in the php file. I would like to add it below the above code, like under the link showed.

<form name="uteam_uploader">
<p align="center">[<span lang="en-us">Click once to copy</span><br>
<input name="linkcode" onclick="Click2Copy('uteam_uploader.linkcode')" value="&quot; . $url_path . &quot;/uploads27/&quot; . $newfilename . &quot;" size="20">
</p>
<p align="center">[<span lang="en-us">Forums use</span><br>
<input name="forumcode" onclick="Click2Copy('uteam_uploader.forumcode')" type="text" value="[img]&quot; . $url_path . &quot;/uploads27/&quot; . $newfilename . &quot;[/img]" size="20"></form>

Code (markup):

Thanks for any help

 

put ?> after the ending braces of if } and paste html code after that and in html to show php variable values use these tags <?= ?> like
<input name="linkcode" onclick="Click2Copy('uteam_uploader.linkcode')" value="&quot; . <?= $url_path ?>. &quot;/uploads27/&quot; . <?= $newfilename ?> . &quot;" size="20">

or simple echo or print variable <? echo $newfilename; ?>

 

hello,

that works but then the form shows on the top of the page, and has all links correctly ,,

but i want to show it under the link in the php code shown above..

 

what i have understand is that you want to show HTML after this line
$uploaderinfo .= $f_link . "<br><a href=\"" . $url_path . "/uploads27/" . $newfilename . "\" target=\"_blank\">" . $url_path . "/uploads27/" . $newfilename . "</a><br><br>\n";

right?

if yes than put ?> after above line and place all html after this tag. Than start php tag <? and close if statement with } ?>

 

mm not exactly,,

you see in this line

what i have understand is that you want to show HTML after this line
$uploaderinfo .= $f_link . "<br><a href=\"" . $url_path . "/uploads27/" . $newfilename . "\" target=\"_blank\">" . $url_path . "/uploads27/" . $newfilename . "</a><br><br>\n";

it shows a link correct,

i would like to put the html after here

$newfilename . "</a><br><br>

I tried what you said, but then the form went at the top,,

 

<?
//...........code
#If link is to be shown, show link

        if ($see_url)

        {

          $uploaderinfo .= $f_link . "<br><a href=\"" . $url_path . "/uploads27/" . $newfilename . "\" target=\"_blank\">" . $url_path . "/uploads27/" . $newfilename . "</a><br><br>\n";
?>
<form name="uteam_uploader">
<p align="center">[<span lang="en-us">Click once to copy</span><br>
<input name="linkcode" onclick="Click2Copy('uteam_uploader.linkcode')" value="&quot; . $url_path . &quot;/uploads27/&quot; . $newfilename . &quot;" size="20">
</p>
<p align="center">[<span lang="en-us">Forums use</span><br>
<input name="forumcode" onclick="Click2Copy('uteam_uploader.forumcode')" type="text" value="[img]&quot; . $url_path . &quot;/uploads27/&quot; . $newfilename . &quot;[/img]" size="20"></form>
<?
        }
		// ............more code
?>		

PHP:

Did you try above code and it did not work. please confirm.

I did not see that you are printing line in your if statement.

  $uploaderinfo .=.....

PHP:

where you echo or print $uploaderinfo? it should be before <form... tag

 

hello this is what i have

        #If link is to be shown, show link

        if ($see_url)

        {

      	  $uploaderinfo .= $f_link . "<br><a href=\"" . $url_path . "/uploads27/" . $newfilename . "\" target=\"_blank\">" . $url_path . "/uploads27/" . $newfilename . "</a><br><br>\n";
?>
<form name="uteam_uploader">
<p align="center">[<span lang="en-us">Click once to copy</span><br>
<input name="linkcode" onclick="Click2Copy('uteam_uploader.linkcode')" value="<?= $url_path ?>./uploads27/<?= $newfilename ?> " size="20">
</p>
<p align="center">[<span lang="en-us">Forums use</span><br>
<input name="forumcode" onclick="Click2Copy('uteam_uploader.forumcode')" type="text" value="[img]<?= $url_path ?>/uploads27/<?= $newfilename ?>[/img]" size="20"></form>
<?
        }

        #If entries are to be logged, log them

PHP:

but the form appears on the top after i upload a file.

 

where is echo $uploaderinfo; or print( $uploaderinfo); statement. You must do it in if condition.

if ($see_url)

        {

          $uploaderinfo .= $f_link . "<br><a href=\"" . $url_path . "/uploads27/" . $newfilename . "\" target=\"_blank\">" . $url_path . "/uploads27/" . $newfilename . "</a><br><br>\n";
       echo $uploaderinfo; // print link here 
?>
html code here

PHP: