How to update a php variable inside a variable

Question

I have the following code:

$var1 = 'Hello';
$var2 = $var1.' friend';

Inside the if statement, if submit is clicked, I want to update var1 to 'Hi' so I am expecting 'Hi friend' as output. But it was not working- 'Hello friend' is still the output. I want to call it via var2. How can var1 be updated which is in var2?

Here is my full code:

<?php 
$var1 = 'Hello';
$var2 = $var1.' friend';

if(isset($_POST['submit'])) {
    $var1 = 'Hi';
    echo $var2;

} else{ 
echo '<form method="post" action="'. $_SERVER['PHP_SELF'] .'">
<input type="text" name="name"><br>
<input type="submit" name="submit" value="Submit Form"><br> </form>';
} 
?>

you must repeat the code .. $var1 = 'Hi'; echo $var1 . $var2; once you have assigned $var1.' friend'; to $var2 .. the value is fixed (assigned ) to $var2 .. and if you want a new one you must reassign — ScaisEdge
– ScaisEdge, Commented Mar 29, 2018 at 16:26
Maybe you should use a function for that. You can't do it this way — Dimitris Filippou
– Dimitris Filippou, Commented Mar 29, 2018 at 16:28

arkascha · Accepted Answer · 2018-03-29 16:30:19Z

2

No, that won't work, you have a miss conception of how things function. $var2 = $var1.' friend'; will assign a value immediately to ??$var2??, not a formular or something evaluated later. So $var2 contains whatever $var1 holds at the time of the assignment.

What you are looking for is a function:

<?php 

function greetWithPhrase($phrase) {
  return $phrase . ' friend';
}

$var1 = 'Hello';

if(isset($_POST['submit'])) {
    $var1 = 'Hi';
    echo greetWithPhrase($var1);    
} else{ 
  echo '<form method="post" action="'. $_SERVER['PHP_SELF'] .'">
  <input type="text" name="name"><br>
  <input type="submit" name="submit" value="Submit Form"><br> </form>';
}

answered Mar 29, 2018 at 16:30

arkascha

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1 Comment

jumbo_siopao Over a year ago

Makes sense. Thanks for the explanation. Im new to php, coming from a java background :P

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