动态访问PHP数组

I tried to access with $this->$arrDataName[$key] on the element with the key $key from the array $this->$arrDataName. But PHP interpretes that wrong.

我尝试使用$ this - > $ arrDataName中的键$ key在元素*问$ this - > $ arrDataName [$ key]。但PHP解释错误。

I tried it with { } around the $arrDataName to $this->{$arrDataName}[$key], but it doesn't work.

我在$ arrDataName周围尝试使用{}到$ this - > {$ arrDataName} [$ key],但它不起作用。

On php.net I found an advice, but I can't realize it.

在php.net上我找到了一个建议,但我无法实现。

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

为了将变量变量与数组一起使用,您必须解决模糊问题。也就是说,如果你写$$ a [1],那么解析器需要知道你是否想要使用$ a [1]作为变量,或者你想要将$$作为变量,然后是[1]索引来自那个变量。解决这种歧义的语法是:第一种情况为$ {$ a [1]},第二种情况为$ {$ a} [1]。

Perhaps anyone can help me.

也许任何人都可以帮助我。

Thanks!

EDIT:

I think it doesn't work, but I forgot to fill the array.
Finally it works. :)
This is the solution: $this->{$arrDataName}[$key]

我认为它不起作用,但我忘了填充阵列。最后它有效。 :)这是解决方案:$ this - > {$ arrDataName} [$ key]

4 个解决方案

#1

Your syntax is correct:

你的语法是正确的:

$this->{$varName}[$key]

You can also use an extra variable for this:

您还可以使用额外的变量:

$myTempArr = $this->$arrDataName;

$myTempArr[ $key ];

IMHO, readability is better that way...

恕我直言,可读性更好......

#2

<?php
    class Foo {
        public function __construct() {
            $this->myArray = array('FooBar');
            $arrayName = 'myArray';
            echo $this->{$arrayName}[0];
        }
    }
    new Foo;

This worked perfectly for me, it printed FooBar.

这对我来说很完美,它印有FooBar。

#3

Let's assume your array is $this->arrDataName. You have a $key, so your object would be $this->arrDataName[$key].

我们假设你的数组是$ this-> arrDataName。你有一个$ key,所以你的对象是$ this-> arrDataName [$ key]。

If you want the contents of the variable which name is stored in $this->arrDataName[$key] you should do this:

如果你想要名称存储在$ this-> arrDataName [$ key]中的变量的内容,你应该这样做:

<?php
    echo ${$this->arrDataName[$key]};
?>

#4

Well, as far as I know, it works. Here how I tested it:

嗯,据我所知,它的确有效。我在这里测试它:

<?php
class tis
{
    var $a = array('a', 'b', 'c');
    var $b = array('x', 'y', 'z');

    public function read($var)
    {
        echo $this->{$var}[1].'<br />';
    }
}

$t = new tis();
$t->read('a');
$t->read('b');
?>

And the output:

并输出:

b
y

Check correctness of $arrDataName. Turn on debuging and displaying PHP erros (including notices). Maybe you're trying to read non-existing property?

检查$ arrDataName的正确性。打开debuging并显示PHP错误(包括通知)。也许你正在尝试阅读不存在的财产?

Also, which PHP version you use? I assume PHP5?

另外,您使用的是哪个PHP版本?我假设PHP5?

#1