Using if(!empty) with multiple variables not in an array

Question

I am trying to polish some code with the if(!empty) function in PHP but I don't know how to apply this to multiple variables when they are not an array (as I had to do previously) so if I have:

$vFoo       = $item[1]; 
$vSomeValue = $item[2]; 
$vAnother   = $item[3];

Then I want to print the result only if there is a value. This works for one variable so you have:

 if (!empty($vFoo)) {
     $result .= "<li>$vFoo</li>";
 }

I tried something along the lines of

if(!empty($vFoo,$vSomeValue,$vAnother) {
    $result .= "<li>$vFoo</li>"
    $result .= "<li>$vSomeValue</li>"
    $result .= "<li>$vAnother</li>"
}

But of course, it doesn't work.

imkingdavid · Answer 1 · 2013-10-11T02:10:50.877

39

You could make a new wrapper function that accepts multiple arguments and passes each through empty(). It would work similar to isset(), returning true only if all arguments are empty, and returning false when it reaches the first non-empty argument. Here's what I came up with, and it worked in my tests.

function mempty()
{
    foreach(func_get_args() as $arg)
        if(empty($arg))
            continue;
        else
            return false;
    return true;
}

Side note: The leading "m" in "mempty" stands for "multiple". You can call it whatever you want, but that seemed like the shortest/easiest way to name it. Besides... it's fun to say. :)

Update 10/10/13: I should probably add that unlike empty() or isset(), this mempty() function will cry bloody murder if you pass it an undefined variable or a non-existent array index.

edited Oct 11 '13 at 02:10

answered Oct 17 '11 at 19:57

imkingdavid

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2

This should be the best answer. A succinct helper function I can reuse in multiple places. – Aditya M P Feb 09 '13 at 06:25
1

Does this not still trigger errors if a variable is undefined, though? – BeesonBison Aug 08 '14 at 20:01
Yes, see my Update at the bottom of the answer; if you pass an undefined variable or non-existent array index, it will display an error. – imkingdavid Aug 08 '14 at 20:12
See: http://sandbox.onlinephpfunctions.com/code/cb35e4e8460a560eca4eb5e60960952df1e88520 and click execute and you will see an error – imkingdavid Aug 08 '14 at 20:14
@fdrv you would call it like this: `$var3 = mempty($var1, $var2);`. If either $var1 or $var 2 has a non-empty value it will return false, otherwise true. – imkingdavid Feb 13 '18 at 03:53

score 21 · Answer 2 · answered Feb 14 '11 at 14:18

You need to write a condition chain. Use && to test multiple variables, with each its own empty() test:

if (!empty($vFoo) && !empty($vSomeValue) && !empty($vAnother)) {

But you probably want to split it up into three ifs, so you can apply the extra text individually:

if (!empty($vFoo)) {
   $result .= "<li>$vFoo</li>";
}
if (!empty($vSomeValue)) {
   $result .= "<li>$vSomeValue</li>";
}
if (!empty($vAnother)) {

score 9 · Answer 3 · answered Feb 14 '11 at 14:18

empty() can only accept one argument. isset(), on the other hand, can accept multiple; it will return true if and only if all of the arguments are set. However, that only checks if they're set, not if they're empty, so if you need to specifically rule out blank strings then you'll have to do what kelloti suggests.

score 7 · Answer 4 · edited Feb 14 '11 at 14:19

7

use boolean/logical operators:

if (!empty($vFoo) && !empty($vSomeValue) && !empty($vAnother)) {
    $result .= "<li>$vFoo</li>"
    ...
}

Also, you might want to join these with or instead of and. As you can see, this can give you quite a bit of flexibility.

edited Feb 14 '11 at 14:19

Phill Pafford

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answered Feb 14 '11 at 14:15

kelloti

8,705
5
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82

score 4 · Answer 5 · answered Jun 01 '16 at 15:58

I use this function as an alternative to isset. It's a little simpler than @imkingdavid's one and return true, if all arguments (variables) are not empty, or return false after first empty value.

function isNotEmpty() {
    foreach (func_get_args() as $val) {
        if (empty($val)) {
            return false;
        }
    }

    return true;
}

score 4 · Answer 6 · answered Jul 19 '17 at 03:06

Just adding to the post by imkingdavid;

To return TRUE if any of the parameters are empty, check out the below instead.

function mempty(...$arguments)
{
    foreach($arguments as $argument) {
        if(empty($argument)) {
            return true;
        }
    }
    return false;
}

This is closer to how isset() works, and is how i would expect empty() to work if supporting a variable length of parameters. This will also keep in line with how empty() works in that it will return FALSE if everything is not empty.

If multiple parameters are supplied then isset() will return TRUE only if all of the parameters are set. Evaluation goes from left to right and stops as soon as an unset variable is encountered.

PHP: isset - Manual

Darn. I didn't see [this post](https://stackoverflow.com/a/37573583/8328731) by [hovado](https://stackoverflow.com/users/2167021/hovado). My apologies. — Dylan, Jul 19 '17 at 03:09

awm · Answer 7 · 2011-02-14T14:56:54.157

3

Save yourself some typing and put it into a loop...

foreach (array('vFoo','vSomeValue','vAnother') as $varname) {
  if (!empty($$varname)) $result .= '<li>'.$$varname.'</li>';
}

edited Feb 14 '11 at 14:56

answered Feb 14 '11 at 14:20

awm

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Isn't that simply wrong code? Wouldn't you do `foreach (array(...) AS $var)` and skip at least one of the variable variables? And anyway, I don't think this 'in' syntax exists in PHP. – Andrew Feb 14 '11 at 14:54
+1 Absolutely right. (What was I thinking? Javascript? Ugh.) Fixed it! I'm confident about the variable variables, though. – awm Feb 14 '11 at 14:57
there is till a problem in this coe.....you need to remove the extra $ in the $varname – pasignature Jul 27 '20 at 13:26

score 2 · Answer 8 · answered Feb 14 '11 at 14:22

2

As others noted, empty() takes only one argument so you have to use something like

if(!empty($v1) && !(empty($v2) ...)

but if (!empty($v)) is the same thing as if($v) so you may also use:

if ($v1 && $v2 ...)

answered Feb 14 '11 at 14:22

Eelvex

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`if (!empty($v))` is **not** the same thing as `if ($v)`. `$v` generates an error if $v is undefined, whereas `empty($v)` returns true if $v is undefined **or** evaluates to (boolean) false. – awm Feb 14 '11 at 14:36
@awn: it is according to the manual, except that one does not generate a *warning*. – Eelvex Feb 14 '11 at 14:37
1

You're right... warning, not error. Still, warnings mess up my page layout. – awm Feb 14 '11 at 14:42
1

@awm: It's not even a warning. It's a notice; semantically a debug message. It indeed depends on the context if syntactic salt for notice suppression is necessary. – mario Feb 14 '11 at 15:26
It's not even a warning, and it **still** messes up my page layout? Darn that `error_reporting(E_ALL)`. – awm Feb 14 '11 at 15:35
check your code `if(!empty($v1) && !empty($v2) ...)` – NineCattoRules Feb 28 '16 at 17:56

Phill Pafford · Answer 9 · 2011-02-14T15:08:07.850

1

Why not just loop through the $item array

foreach($item as $v) {
    if(!empty($v))) {
        $result .= "<li>$v</li>";
    }
}

You could also validate of the key index value as well

foreach($item as $key => $value) {
    if($key == 'vFoo') {
        if(!empty($value))) {
            $result .= "<li>$value</li>";
        }
    }
    // would add other keys to check here
}

For the empty() versus isset()

$zero = array(0, "0");

foreach($zero as $z) {
    echo "\nempty(): $z ";
    var_dump($z);
    var_dump(empty($z)); // Will return true as in it's empty
    echo "isset(): $z ";
    var_dump(isset($z)); // will return true as it has value
}

Output:

empty(): 0 int(0)
bool(true)
isset(): 0 bool(true)

empty(): 0 string(1) "0"
bool(true)
isset(): 0 bool(true)

edited Feb 14 '11 at 15:08

answered Feb 14 '11 at 14:24

Phill Pafford

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What's the point of `isset($v)` here? We already know `$v` is set because it's the parameter of the foreach loop. – awm Feb 14 '11 at 14:31
Pendantic note, the `isset` call is redundent here, since there is no case where `!empty($v)` will be true where `isset($v)` will be false. So that simplifies to `!empty($v)`. Now, if it wasn't in a loop, you should put the `isset()` call first since the variable may not be set. But in the code you provided, just get rid of it all together... – ircmaxell Feb 14 '11 at 14:31
I like to use the isset() as well to check for 0 values but I had it as && instead of ||. – Phill Pafford Feb 14 '11 at 14:34
@Phill: Hi Phill, I'm not sure I can pass them to one array as each variable has it's own name + value i.e.
$Foo: $ValueofFoo

HGB

Feb 14 '11 at 14:54

@Nathan: I would have to print them only if they have a value, but if either of them is empty the others still need to be printed. – HGB Feb 14 '11 at 14:57

@Cigana Look at your own code! `$vFoo = $item[1]; $vSomeValue = $item[2]; $vAnother = $item[3];` Why do you have to use the named variables? They come directly out of the `$item` array; that's why Phill wrote this! – awm Feb 14 '11 at 15:03

@awm see my empty() versus isset() example – Phill Pafford Feb 14 '11 at 15:03

@Everyone - And there is also a new change... is not just going to be the same output result for all so one can be

$variable

and the other can be

$variableSome harcoded text here

,

$variable

...Sigh... things can never be simple! :( – HGB Feb 14 '11 at 15:04

score 0 · Answer 10 · answered Jun 13 '17 at 16:34

0

Why not just like this:

foreach(array_filter($input_array) as $key => $value) {
    // do something with $key and $value
}

answered Jun 13 '17 at 16:34

dutch

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1
5

score 0 · Answer 11 · answered Nov 16 '17 at 13:53

0

Just for the record: array_reduce works as well.

if (!array_reduce($item, function($prev_result,$next) {
    return $prev_result || empty($next); // $prev_result will be initial null=false
})) {
    // do your stuff
}

answered Nov 16 '17 at 13:53

Liam

1

score 0 · Answer 12 · answered Mar 04 '18 at 14:58

0

If working only with strings, you may try the following:

if(empty($str1.$str2.$str3)){ /* ... */ }

Note, that empty(0) would return true, while empty('0'.'0') and empty('0') won't.

answered Mar 04 '18 at 14:58

Leo

1,508
13
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1

I should had mentioned that this will not work that way the op asked because the equal function will return true if all strings are empty in this case. You can not use this method to validate if all strings are NOT empty, only if not all are empty: `!empty($str1.$str2)`is equal to `!empty($str1) or !empty($str2)` wich is not the same as `!empty($str1) and !empty($str2)`. – Leo Mar 15 '18 at 20:29
But it's an easy and short hand way to proove if all strings are empty. – Leo Mar 15 '18 at 20:36
why only with string? – Alex78191 Dec 20 '21 at 11:59
Well, back then variables in php were not typed (not saying they are today, but it's possible now to somewhat enforce types). Also, how should the concept of emptyness be translated to other types i.e. numbers or boolenas? – Leo Dec 21 '21 at 11:45
variables can contain `null`, `0` or `false` values. they are all treated as empty – Alex78191 Dec 21 '21 at 11:54
That is true, but not when concatenated. E.g. the number zero when concatenated will be tretaed as a string, more precisely the character '0'. https://www.php.net/manual/de/language.operators.string.php – Leo Dec 21 '21 at 12:26

score 0 · Answer 13 · answered Feb 14 '11 at 14:20

I'm not 100% sure I understand what you're trying to do, but here are a couple of possible answers:

This will return each variable only if it isn't empty:

function printIfNotEmpty($var) {
  if (!empty($var)) {
    return '<li>' . $var . '</var>';
  }
  return '';
}
$result .= printIfNotEmpty($vFoo);
$result .= printIfNotEmpty($vSomeValue);
$result .= printIfNotEmpty($vAnother);

Or this one will add all three of them if none of them are empty:

if (!empty($vFoo) && !empty($vSomeValue) && !empty($vAnother)) {
  $result .= '<li>' . $vFoo . '</li>';
  $result .= '<li>' . $vSomeValue . '</li>';
  $result .= '<li>' . $vAnother . '</li>';
}

shouldn't it be written like this `if (!empty($vFoo && $vSomeValue && $vAnother) )`? — Learning, Jun 20 '17 at 09:24

score 0 · Answer 14 · answered Feb 15 '11 at 11:53

Guys, many thanks for all your responses, I kind of created a script based on all of our possible answers which is actually helping me with learning PHP's syntax which is the cause behind most of my errors :) So here it is

$vNArray ['Brandon']  = $item[3]; 
$vNArray['Smith']= $item[4]; 
$vNArray ['Johnson']= $item[5];
$vNArray ['Murphy']= $item[6];
$vNArray ['Lepsky']= $item[7];

foreach ($vNArray as $key => $value){

    if(!empty($value)){
        $result  .= "\t\t\t\t<li><strong>$key</strong>"  .$value.   "</li>\n";
}

I just need to figure out how to make some of these different now. So could I access each in the following way?

$result  .= "\t\t\t\t<li><strong>$key[0]</strong>"  .$value. "</li>\n";
$result  .= "\t\t\t\t<li><a href="thislink.com">$key[3]</a>"  .$value. "</li>\n";

Thanks guys!

score 0 · Answer 15 · answered Nov 25 '21 at 05:09

0

You can try php &&/ and operator. They both are same.

if(!empty($vFoo) && !empty($vSomeValue) && !empty($vAnother) {
   $result .= "<li>$vFoo</li>"
   $result .= "<li>$vSomeValue</li>"
   $result .= "<li>$vAnother</li>"
}

answered Nov 25 '21 at 05:09

Mohammudullaha Khan

3
1
2

Using if(!empty) with multiple variables not in an array

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