Php 动态变量

Question

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

这是有效的：

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

但这不起作用：

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

为什么？我怎样才能让它发挥作用？

Answer 1

您应该向 php 解析器解释您希望如何解析此语句：

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

没有括号你将有：

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

Sample link.