combining two php variables

Question

    $current_month  = date('M'); 
    for($m=1; $m<=$current_month; ++$m)
    {
        $monthNamelower=strtolower(date('M', mktime(0, 0, 0, $m, 1))).'';

            if ($jandata!='')
            {
              echo $jandata;
            }

        echo $data2jan;  
    }

can i replace the month jan with the variable $monthNamelower?
I tried something like this but not working

    if ('$'.$monthNamelower.'data'!='')
    {
            echo '$'.$monthNamelower.'data'; 

    }
    echo '$data2'.$monthNamelower;

It is quite difficult to understand your question, maybe it is better to edit it stackoverflow.com/help/how-to-ask. From where pops $jandata out? Maybe you would better use an array to store your data without using variables of variables (and if your data are already in place compact is your best friend!) — Eineki
– Eineki, Commented Jul 5, 2017 at 7:01

unalignedmemoryaccess · Accepted Answer · 2017-07-05 06:45:19Z

3

Yes, you can do it.

if (${$monthNamelower . 'data'} != '') {
    echo ${$monthNamelower . 'data'}; 
}

If your value stored in $monthNamelower is j, it will check for variable name jdata and will print it.

You can test this code directly to see effect:

$monthNamelower = 'j';
$jdata = 'January';
if (${$monthNamelower . 'data'} != '') {
    echo ${$monthNamelower . 'data'};  //Prints 'January'
}

answered Jul 5, 2017 at 6:45

unalignedmemoryaccess

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1 Comment

gtroop Over a year ago

thank you ..clear explanation but not sure why its not working in my code ..my old variable is working..i'll try to figure it out first

Bibhudatta Sahoo · Accepted Answer · 2017-07-05 06:51:42Z

1

You can do this by making your variable name as dynamic Just like this

$monthNamelower.='data';
if ($$t!='')
    {
echo $$t; 

}

For Example:-

$t='jan';
$jandate='good';
$t.='date';
echo $$t;

It will produce output like

I thing it will help you.

answered Jul 5, 2017 at 6:51

Bibhudatta Sahoo

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