PHP Use variable name in foreach after setting value

Question

EDIT: SOLVED

ORIGINAL POST:

I have a foreach loop which creates variables in the form of:

    foreach ($fieldsarray as $field)
    {
        ${$field} = $globalrow[$field];
    }

This creates variables using array elements, like $firstname, $lastname, etc.

Now, since the values of these variables are fetched from the DB, I cannot use the variable names again in the same foreach. What I would like to do is to use this in the loop:

if(!empty($$field)) 
  {
    $r.$field='<tr><td width="50%">'.${$field}._label.': </td>
    <td width="50%">'.${$field}.'</td></tr>';
  }

Basically, this would generate table rows with data from DB, but only if the value of the variable is not empty.

My problem is: Since I have declared above that e.g. $firstname = 'John' - then the second part of the code does not work anymore, because I assume PHP does not transform the above snippet into

if(!empty($firstname))

but rather into

if(!empty(John))

How can I "force" PHP to again use the variable NAME rather than its value in this case?

Thank you for your help.

EDIT: OK, cheezeburgers suggestions worked and I think now it's working as intended. Thanks to all of you for your help.

EDIT 2: The table rows are output correctly on the screen, e.g.:

First Name: John
Last Name : Smith

but for some weird reason in the email confirmation the output is doubled:

First Name: John
Last Name : Smith
First Name: John
Last Name : Smith

In both cases I simply echo them with:

<?php echo $allrows ?>

where $allrows is in the foreach loop.

(I will post this part as a new question since this seems to be a new problem now)

Urgh, that gives me a headache, variable variables are not easy to read. What are you actually trying to achieve? Im positive there is a much better way — Steve
Why do you need the variable variables? Can't you use the array for table generation? (hint: you can) — kapa
But I think it should work as you wrote it. PHP does transform empty($$field) into empty($firstname) when $field = 'firstname'. — Barmar
Why on earth do you want to do this? You already have a variable (the array) with all the info you need. This makes no sense to me — PeeHaa

AyB AyB · Accepted Answer · 2014-04-24T11:21:00

PHP will work it out fine, just that you have missed the quote around _label and probably that's being read as a constant.

Try this:

$$field='<tr><td width="50%">'.${$field}.'_label: </td>
<td width="50%">'.${$field}.'</td></tr>';

which gives the output as:

John_label: John

Demo

PHP Use variable name in foreach after setting value

2 Answers