Construct a PHP variable name based on other variable values and static text

Question

I want to tell my function which variable to call based on the day of the week. The day of the week is stored in $s_day, and the variables I want to call changes based on which day it is.

e.g.

I've stored a string 'Welcome to the week' in $d_monday_text1. Rather than build a set of 7 conditional statements (e.g. if date=monday echo $foo, else if date=tuesday echo $bar...), can I change the name of the variable called in the function by concatenating the name of the variable?

$s_day = date("l");
$text1 = '$d_'.$s_day.'_text1';

I'm hoping this evaluates to $d_monday_text1, which, as mentioned above, has the value "Welcome to the week". So, later on I'd want to use:

echo $text1;

To yield the resulting output = Welcome to the week.

I've looked into variable variables, which may be the way to go here, but am struggling with syntax. I can get it to echo the concatenated name, but I can't figure out how to get that name evaluated.

possible duplicate of Can I use a generated variable name in PHP? — outis
– outis, Commented Oct 1, 2010 at 4:04

Jani Hartikainen · Accepted Answer · 2009-11-30 05:10:00Z

10

Variable variables aren't a good idea - You should rather use arrays. They suit this problem much, much better.

For example, you could use something like this:

$messages = array(
    'monday' => 'Welcome to the week',
    'tuesday' => 'Blah blah',
    'wednesday' => 'wed',
    'thursday' => 'thu',
    'friday' => 'fri',
    'saturday' => 'sat',
    'sunday' => 'week is over!'
);

$dayName = date('l');
echo $messages[$dayName];

Arrays are the data format used to store multiple related values such as these.

answered Nov 30, 2009 at 5:10

Jani Hartikainen

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3 Comments

jkramp Over a year ago

Thanks Jani. I think this would work for a smaller dataset than I have. It does fit the example given, which is a simplified explanation of what I'm doing here.

Jani Hartikainen Over a year ago

In general I think arrays should be used whenever you have a dataset of any size. Using dynamic naming for variables is generally a poor idea, as it does not give the data any structure and it makes the code less readable.

jkramp Over a year ago

Okay, let me noodle on it a bit more and I'll update the question to see if there's a better way to go using arrays. Probably due to the way my data is structured it is easier to use the dynamic expression than the array. that's not to say my data structure isn't flawed.

Kenji Kina · Accepted Answer · 2009-12-11 06:28:10Z

4

You can use this syntax:

$$text1

I've used this before. The evaluation comes as in:

$($text1)

edited Dec 11, 2009 at 6:28

answered Nov 30, 2009 at 5:05

Kenji Kina

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Comments

Sergey Onishchenko · Accepted Answer · 2014-07-01 14:21:08Z

0

Let's consider the following example:

$construction_no = 5;
$construction_5 = 'Door';

$var_name = 'construction_'.$construction_no;
$var_value = ${$var_name};
echo $var_value; //Door

Object oriented approach

$var_value = $this->{$var_name};
echo $var_value; //Door

answered Jul 1, 2014 at 14:21

Sergey Onishchenko

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