PHP use variables in an array

Question

Currently I am setting an array up like this...

public function setup_array()
{
    $columns = array(
        'date1' => '2014-01-24',
        'date2' => '2014-02-14',
        'date3' => '2014-03-11',
        'date4' => '2014-04-01'
    );
    return $columns;
}

This works but what I would like to use variables that are already set in place of the dates like so...

public function setup_array()
{
    $columns = array(
        'date1' => '$date1',
        'date2' => '$date1',
        'date3' => '$date1',
        'date4' => '$date1'
    );
    return $columns;
}

I have tried to do this an although the $date variables are avaliable and set, it actually prints $date1 instead or retrieving the variable value itself.

What am I doing wrong?

Your $date1 variable will need to be accessible from within the scope of the function — Mark Baker
– Mark Baker, Commented Apr 21, 2014 at 13:25

Alma Do · Accepted Answer · 2014-04-21 13:25:52Z

4

Single quotes will cause to interpret string as it is, literally. use double-quotes.

Without any passed parameter - $date1 is unknown inside function scope and, thus, would not be substituted.

public function setup_array($date1)
{
    $columns = array(
        'date1' => $date1,
        'date2' => $date1,
        'date3' => $date1,
        'date4' => $date1
    );
    return $columns;
}

In your case even double quotes aren't needed since you're using variable as it is.

answered Apr 21, 2014 at 13:25

Alma Do

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Comments

Lucas Henrique · Accepted Answer · 2014-04-21 13:26:15Z

2

Remove quotes

public function setup_array($date1)
{
    $columns = array(
        'date1' => $date1,
        'date2' => $date1,
        'date3' => $date1,
        'date4' => $date1
    );
    return $columns;
}

answered Apr 21, 2014 at 13:26

Lucas Henrique

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Comments

Burhan Çetin · Accepted Answer · 2014-04-21 13:25:30Z

1

Remove quotes

    public function setup_array($date1)
{
    $columns = array(
        'date1' => $date1,
        'date2' => $date1,
        'date3' => $date1,
        'date4' => $date1
    );
    return $columns;
}

answered Apr 21, 2014 at 13:25

Burhan Çetin

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1 Comment

John Conde Over a year ago

You should explain what you changed and why you did it

Your Common Sense · Accepted Answer · 2014-04-21 13:27:28Z

0

You need to understand PHP syntax

Quotes in your first example don't mean "something that is used as array value" but regular string delimiters. I.e. these quotes belong to strings, not to array.

answered Apr 21, 2014 at 13:27

Your Common Sense

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Comments

Manwal · Accepted Answer · 2014-04-21 13:32:21Z

0

Remove Single quotes

public function setup_array()
{
    $columns = array(
        'date1' => $date1,
        'date2' => $date1,
        'date3' => $date1,
        'date4' => $date1
    );
    return $columns;
}

Explanation:

$expand=1;$either=2;
echo 'Variables do not $expand $either';
// Outputs: Variables do not $expand $either

echo "Variables do not $expand $either";
// Outputs: Variables do not 2 1

More Details String PHP

edited Apr 21, 2014 at 13:32

answered Apr 21, 2014 at 13:26

Manwal

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2 Comments

Your Common Sense Over a year ago

...and add double. hilarious

juanmf Over a year ago

Mmm doing the same thing twice expecting different results.. I know a definition for that :P

juanmf · Accepted Answer · 2014-04-21 13:45:54Z

0

Also note that \DateTime does not have a __toString() method, so if $date1 is an instance of \DateTime, you must call $date1->format('Y-m-d')

public function setup_array($date1)
{
    $columns = array(
        'date1' => $date1->format('Y-m-d'),
        'date2' => $date1->format('Y-m-d'),
        'date3' => $date1->format('Y-m-d'),
        'date4' => $date1->format('Y-m-d'),
    );
    return $columns;
}

otherwise it'd throw an \Exception in the event of echoing or casting to string.. if $date1 is scalar you can use double quotes "$date1" and it should still work. See [http://www.php.net/manual/es/class.datetime.php]

edited Apr 21, 2014 at 13:45

answered Apr 21, 2014 at 13:29

juanmf

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