php if statement within/inside single variable?

Question

I'm not sure if this will work, or how to make it work.

Any help would be so great thanks.

I need to make this contitional statement string into one variabe?

How can I do this?

if ($address1) {
    echo $address1 . ', ';
}

if ($address2) {
    echo $address2 . ', ';
}

if ($address3) {
    echo $address3 . ', ';
}

if ($address4) {
    echo $address4 . ', ';
}

if ($townCity) {
    echo $townCity . ', ';
}

if ($county) {
    echo $county . ', ';
}

if ($postCode) {
    echo $postCode;
}

So basically I need this in one variable, and the reason I have it in if statements, is because it is possible for the some off these variables to not exist.

And after each variable, it has a comma and space.

If I do it like this...

$singleVar =    $address1 . ', ' . 
                $address2 . ', ' .
                $address3 . ', ' . 
                $address4 . ', ' . 
                $townCity . ', ' . 
                $county . ', ' . 
                $postCode;

Then for the variables that do not exist, will have and extra comma and space - not cool.

Can any one please advise.

Thanks

you might want to have a look at php.net/manual/en/language.operators.comparison.php -> Ternary Operator — Rudolf Mühlbauer
– Rudolf Mühlbauer, Commented Oct 16, 2012 at 21:13
You need this i think php.net/manual/en/language.operators.comparison.php#example-120 — rsz
– rsz, Commented Oct 16, 2012 at 21:15

No Results Found · Accepted Answer · 2012-10-16 21:30:22Z

7

Lots of ways to do it, but here's what I'd use:

$items = array(
    $address1,
    $address2,
    $address3,
    $address4,
    $townCity,
    $county,
    $postCode
);
echo implode(', ', array_filter($items));

array_filter removes the "empty" ones, see converting to boolean in the PHP docs.

implode joins the array items together by the string you pass as the first argument, and makes sure there is no extra , dangling off the end of the output.

If you're worried the variables may not defined, use something like this instead:

$items = array();
$fields = array(
    'address1',
    'address2',
    'address3',
    'address4',
    'townCity',
    'county',
    'postCode'
);
foreach ($fields as $field)
{
    isset($$field) && $items[] = $$field;
}
echo implode(', ', $items);

Just remember that isset checks if something is "set", not if it is "empty". For example, a variable with an empty string or a zero is "set", but fails the if ($var) condition.

edited Oct 16, 2012 at 21:30

answered Oct 16, 2012 at 21:14

No Results Found

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9 Comments

GolezTrol Over a year ago

Nice solutions, if the variables indeed exist. If not, you'll need an isset somewhere.

Nelson Benítez León Over a year ago

Yes, this will fail when any of those variables does not exist.. which is the reason the original OP code had so many IF's..

No Results Found Over a year ago

@GolezTrol: array_filter will remove them, but yes I agree.

No Results Found Over a year ago

@Nelson: The OP is not using isset, but if ($var) which is much different.

Joshc Over a year ago

Thank @Welsey and for elaborating. Will try this. Can I ask, will there be a ', ' after the final variable?

|

kodeart · Accepted Answer · 2012-10-16 21:33:53Z

1

$address1 = 'foo';
$address3 = 'bar';
$postCode = 'baz';

$vars = array();
foreach (array('address1', 'address2', 'address3', 'address4', 'townCity', 'county', 'postCode') as $var)
{
    isset($$var) and $vars[] = $$var;
}
$single = implode(', ', $vars);

Test http://codepad.org/WwzRYvuO

answered Oct 16, 2012 at 21:33

kodeart

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Comments

GolezTrol · Accepted Answer · 2012-10-16 21:14:17Z

0

If the variables actually don't exist, you'll need isset to check for that. If they do exist but are empty/null/false, you may use the answer by Wesley Murch.

$singleVar = '';
$variables = array('address1','address2','address3',
                   'address4','townCity','county','postCode');

foreach($variables as $variable)
{
    if (isset($$variable))
      $singleVar .= ($$variable . ', ');
}

answered Oct 16, 2012 at 21:14

GolezTrol

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5 Comments

No Results Found Over a year ago

I can only guess the problem is the trailing , (not my dv).

GolezTrol Over a year ago

You are right. It would be easy to trim it, or use this loop to build a new array, which can then be imploded. This method shows the use of variable variables combined with isset, but if the code posted by OP works, the variables actually do exist and this answer is obsolete.

No Results Found Over a year ago

My only fear with trim is if one of the actual values contains a character that you would trim.

GolezTrol Over a year ago

I meant, cutting it off using substr or something, not actually using trim. (sorry). :)

Joshc Over a year ago

Thank for you method too, will try. Interesting reading your comment.

Nelson Benítez León · Accepted Answer · 2012-10-16 21:17:17Z

0

You can do it using variable variables, like so:

$arr = array('address1','address2','address3','address4','townCity','county','postCode');
$singleVar = '';
foreach ($arr as $val) {
   if (isset($$val)) {
      $singleVar .= $$val . ', ';
   }
}

answered Oct 16, 2012 at 21:17

Nelson Benítez León

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Comments

Community · Accepted Answer · 2017-05-23 12:12:26Z

-1

i would propose the following:

join(array_filter(array($a,$b,$c,$d,$e,$f)), ", ");

with $a, etc being your variables. cf. https://stackoverflow.com/a/2603047/1689451

edited May 23, 2017 at 12:12

CommunityBot

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answered Oct 16, 2012 at 21:23

Rudolf Mühlbauer

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Comments

Landon · Accepted Answer · 2012-10-17 16:50:38Z

-1

$string='';
if($address1)
  $string.= $address1.',';
if(address2)
  $string.= $address2.',';
//...etc
echo $string;

edited Oct 17, 2012 at 16:50

answered Oct 16, 2012 at 21:14

Landon

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2 Comments

Landon Over a year ago

hmmm, why the downvote with no comment? this solution works fine

Andrew Barber Over a year ago

Perhaps due to lack of explanation?

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