Php Dynamic Variable

Question

$X['high'] = 1234;
$var = array("X","high");

This is working:

$temp = $$var[0];
$temp = $temp[$var[1]];
echo $temp;

But this isn't working:

echo $$var[0][$var[1]];

Why? How can i make it works?

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

You should explain to php parser how you want this statement to be parsed:

echo ${$var[0]}[$var[1]];

Without brackets you will have:

php7

Notice: Array to string conversion in /in/cvZqc on line 5

Notice: Undefined variable: Array in /in/cvZqc on line 5

php5

Warning: Illegal string offset 'high' in /in/cvZqc on line 5

Notice: Array to string conversion in /in/cvZqc on line 5

CommunityBot

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answered Dec 22, 2016 at 12:58

u_mulder

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1 Comment

Make it perfect and show , before and after PHP7 syntax, or just notice that :)