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PHP Notice: Array to string conversion only on PHP 7

Tags:

php

php-7

I am a newbie of PHP. I study it from php.net, but I found a problem today.

class foo {
    var $bar = 'I am bar.';
}

$foo = new foo();
$bar = 'bar';
$baz = array('foo', 'bar', 'baz', 'quux');
echo "{$foo->$bar}\n";
echo "{$foo->$baz[1]}\n";

The documentation(http://php.net/manual/en/language.types.string.php) say that the above example will output:

I am bar.
I am bar.

But I get the different output run on my PC(PHP 7):

I am bar.
<b>Notice</b>:  Array to string conversion in ... on line <b>9</b><br />
<b>Notice</b>:  Undefined property: foo::$Array in ... on line <b>9</b><br />

Why?

like image 282
lizs Avatar asked Mar 10 '16 02:03

lizs


1 Answers

This should work with PHP 7:

class foo {
var $bar = 'I am bar.';
}

$foo = new foo();
$bar = 'bar';
$baz = array('foo', 'bar', 'baz', 'quux');
echo "{$foo->$bar}\n";
echo "{$foo->{$baz[1]}}\n";

This is caused because in PHP 5 the following line:

echo "{$foo->$baz[1]}\n";

is interpreted as:

echo "{$foo->{$baz[1]}}\n";

While in PHP 7 it's interpreted as:

echo "{{$foo->$baz}[1]}\n";

And so in PHP 7 it's passing the entire array to $foo instead of just that element.

like image 74
C.Liddell Avatar answered Oct 29 '22 10:10

C.Liddell