double $ (variable of variable) doesn't print the variable value as a variable

Question

I have the following code:

<?php
$dates = array('2014-12-01','2014-12-02','2014-12-08','2014-12-09','2014-12-10','2014-12-11');
$values = array(5,3,7,8,9,2);

foreach ($dates as $date){
 //Array of the regarded days names is generated
  $days[] = strtolower(date('l', strtotime($date)))."\n";
}

for ($i = 0; $i < count($days); $i++){
 $day = $days[$i];    
 $$day = $values[$i];
}    
echo $monday;
?>

echo $monday does not print any value, I expect it to print 8, because in the last loop I had a variable named with the value of $day and the last setting for that value should be 8. So why it does not set correctly?!

This is a demo: http://codepad.org/VDIyBuq3

I don't ask about the regarded question itself. I just ask about other implementation. — SaidbakR
– SaidbakR, Commented Dec 13, 2014 at 0:43

Funk Forty Niner · Accepted Answer · 2014-12-12 23:09:06Z

4

This is your problem:

$days[] = strtolower(date('l', strtotime($date)))."\n";
                                                 ^^^^^ here

You are adding a new-line character to the end of your value, so your value will not be monday but monday\n.

Just remove that:

$days[] = strtolower(date('l', strtotime($date)));

edited Dec 12, 2014 at 23:09

Funk Forty Niner

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answered Dec 12, 2014 at 23:06

jeroen

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2 Comments

jeroen Over a year ago

@Fred-ii- A bit bored I guess ;-)

Funk Forty Niner Over a year ago

Hahaha!! You know me ;) Always the perfectionist

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