Combining PHP variables in a while loop

Question

Not sure if the Title is the correct title but I'm not sure how to call what I'm trying to accomplish. On the surface it looks fairly simple...

I wrote the following code:

$var1 = "55";
$var2 = "66";
$var3 = "22";
$var4 = "22";
$x = 1;
$result = "null";

while($x < 5) {
$result = "$" . "var" . "$x";
echo $result;
echo " | ";
$x = $x +1;
}

Was hoping to get: 55 | 66 | 22 |22

Instead I'm getting: $var1 | $var2 | $var3 | $var4 |

I tried many things but unable to find a working solution, any help would be much approached.

Thank you

you are building a string called $var1 when you mean to evaluate that string to the value of $var1. I guess you don't know what variable names you will handle beforehand. However, using eval on an unknown input stream is a bad approach. Use arrays and iterate over them as @PeeHaa said. — ffflabs
– ffflabs, Commented Dec 21, 2016 at 20:14
Array is a great idea but in this case I really would like to find a way to make it work without an array. The code above is a simplified version of what I'm trying to do which does contain four different arrays. I figured it would be easier to post this strip down code that emulate the problem. — badatz
– badatz, Commented Dec 21, 2016 at 20:20

Rajdeep Paul · Accepted Answer · 2016-12-21 22:11:47Z

5

See this statement here,

$result = "$" . "var" . "$x";

$result variable is actually a string. To get it's value, use variable variables. Also, with your approach you'll end up getting an extra | at the end. So to achieve your desired result, change your while() loop in the following way,

while (true) { 
    $result = "var" . $x; 
    echo $$result;
    if($x++ == 4) break;
    echo " | ";
}

Here's a demo

Furthermore, do look into @freginold's solution as well, which I slightly modified to suit your needs. Initialize $x with 5 and change your while() loop in the following way,

$output = "";
while (--$x) { 
    $result = "var" . $x; 
    $output = $$result . " | " . $output; 
}
echo rtrim($output, " | ");

In this solution, while coming out of the loop you'll end up getting an extra | at the end, which you need to trim down using rtrim() function.

Here's a demo

edited Dec 21, 2016 at 22:11

answered Dec 21, 2016 at 20:12

Rajdeep Paul

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5 Comments

freginold Over a year ago

This is the best solution so far. You could even condense your code further by including $output=""; in your initializations and then changing your while loop to this: while (--$x) { $result = "var" . "$x"; $output = $$result . " | " . $output; } with echo $output; at the end.

Rajdeep Paul Over a year ago

@freginold Good attempt,;-) but there's one flaw in your code, which I noticed now. In your method the elements will be printed in reverse order, given that we initialize $x with 5.

freginold Over a year ago

Of course you would also have to change $x = 1; to $x = 5;

freginold Over a year ago

they are processed in reverse order, but printed in correct order, via adding the new values to the beginning of the output string :)

Rajdeep Paul Over a year ago

@freginold Ah, you're right, I didn't think it through. Nice solution! I'll add it in my answer as well. Cheers! :-)

u_mulder · Accepted Answer · 2016-12-21 20:14:54Z

1

Another solution without a while loop:

$var1 = "55";
$var2 = "66";
$var3 = "22";
$var4 = "22";

$vars = array($var1, $var2, $var3, $var4);
echo implode(' | ', $vars);

answered Dec 21, 2016 at 20:14

u_mulder

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Comments

Dominion79 · Accepted Answer · 2016-12-21 20:14:05Z

0

Would an array work better?

<?php 

  $vars = array( 
    'var1' => 55, 
    'var2' => 66, 
    'var3' => 22, 
    'var4' => 22
    'x'    => 1
  );

  foreach($vars as $key => $value){
    print $value;
  }

?>

answered Dec 21, 2016 at 20:14

Dominion79

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