Variable variables created within a for loop

Question

I'd like to change this:

if ($week_day == "1")
{
$day1_hours = $value;
}
if ($week_day == "2")
{
$day2_hours = $value;
}
if ($week_day == "3")
{
$day3_hours = $value;
}
if ($week_day == "4")
{
$day4_hours = $value;
}
if ($week_day == "5")
{
$day5_hours = $value;
}
if ($week_day == "6")
{
$day6_hours = $value;
}
if ($week_day == "7")
{
$day7_hours = $value;
}
}

Into something more readable, like a for loop, or whatever other suggestions you all may have.

I tried to do:

for ($c=1; $c<8, $c++)
{
if ($week_day == $c)
{
$day".$c."_hours = $value;
}
}

But I know that is nowhere near correct, and I have no idea how to insert another variable within a variable.

Any help is appreciated!

Wouldn't it be easier to set an array key? $day[$week_day] = $value — andrewsi
– andrewsi, Commented Jul 3, 2013 at 0:50
Check manual for more info php.net/manual/en/language.types.array.php. Please try your own code. Posting invalid syntax doesn't show an attempt at solving the problem... — elclanrs
– elclanrs, Commented Jul 3, 2013 at 0:52
Could you post the full code for this situation as an answer? Declaring the array and so on? — RobDubya
– RobDubya, Commented Jul 3, 2013 at 0:52

sectus · Accepted Answer · 2013-07-03 00:52:15Z

1

Try this syntax.

${'day'.$c.'_hours'} = $value;

answered Jul 3, 2013 at 0:52

sectus

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1 Comment

RobDubya Over a year ago

No joy, loads a blank page.

Kris · Accepted Answer · 2013-07-03 00:56:26Z

0

Try this

$hours[$week_day] = $value

answered Jul 3, 2013 at 0:56

Kris

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Comments

Lloyd Banks · Accepted Answer · 2013-07-03 01:04:23Z

0

Something like this

$week_day = 3;
$value = "hi";

${"day" . $week_day . "_hours"} = $value;  

echo $day3_hours;

answered Jul 3, 2013 at 1:04

Lloyd Banks

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Drixson Oseña · Accepted Answer · 2013-07-03 01:17:16Z

0

My interpretation.

$week = 7;
$day = array();

for($i=1; $i<=7; $i++){
   $day[$i]['hours'] = $value;
}

You can declare the numbers of weeks you want and to pull the data you can do something like this:

echo $day[1]['hours']

answered Jul 3, 2013 at 1:17

Drixson Oseña

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