PHP注意:仅在PHP 7上将数组转换为字符串 [英] PHP Notice: Array to string conversion only on PHP 7

问题描述

我是PHP的新手.我从php.net进行了研究，但是今天发现了一个问题.

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";

文档( http://php.net/manual/zh/language .types.string.php )表示上面的示例将输出:

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

I am bar.
I am bar.

但是我在PC(PHP 7)上运行了不同的输出:

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 />

为什么?

推荐答案

这适用于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";

这是因为在PHP 5中出现以下行:

This is caused because in PHP 5 the following line:

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

被解释为:

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

在PHP 7中，它解释为:

While in PHP 7 it's interpreted as:

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

因此，在PHP 7中，它将整个数组传递给$foo，而不仅仅是该元素.

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

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