PHP/HTML printing two variables with an underscore inbetween?

Question

Hello StackOverflow Community,
I want to be able to echo that a file is uploaded in upload/FirstName_LastName_FileName.ext format. In my PHP file, I have this HTML here:

<td>File Stored in:</td>
<td>upload/$uploadName_$filename</td>

but it seems that this code returns upload/$filename. How can I remedy this situation? I'm very new to this so help is greatly appreciated!

EDIT: I was too hasty in trying to get this question answered that I didn't provide enough information. I actually could not use the PHP tags since it was going to be embedded into the message of the mail function. So I found a way by assigning a new variable $uploadPath to the code. The line now looks lke this:

   $uploadPath = "upload/" . $uploadName . "_" . $filename;
   // mail to, subject, message and its table...
   <td>File Stored in:</td>
   <td>$uploadPath</td>

Even still, thank you all for your quick answers! I gave the fastest the checkmark.

John Conde · Accepted Answer · 2014-04-01 18:50:38Z

1

There's a lot of ways to do this. Here's a few:

<td>upload/<?= $uploadName,'_',$filename; ?></td>

<td>upload/<?= $uploadName.'_'.$filename; ?></td>

<td>upload/<?php printf("%s_%s", $uploadName, $filename; ?></td>

<td>upload/<?= $uploadName; ?>_<?= $filename; ?></td>

<td>upload/<?= "{$uploadName}_{$filename}"; ?></td>

answered Apr 1, 2014 at 18:50

John Conde

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2 Comments

Federico J. Over a year ago

Advise about php.ini and short_tags!, ;D

John Conde Over a year ago

Yes, if they are using PHP < 5.4 or plan to distribute their code. Otherwise this is fine since in PHP 5.4+ short echo tags are always on. In fact, this is default functionality in all non-deprecated versions of PHP.

Federico J. · Accepted Answer · 2014-04-01 18:51:07Z

0

Use php tags:

<td>File Stored in:</td>
<td>upload/<?php echo $uploadName?>_<?php echo $filename?></td>

answered Apr 1, 2014 at 18:51

Federico J.

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Comments

ponciste · Accepted Answer · 2014-04-01 18:51:43Z

0

try this code:

<td><?php echo 'upload/' . $uploadName . '_' . $filename; ?>**strong text**</td>

answered Apr 1, 2014 at 18:51

ponciste

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