在 PHP 中使用带有动态变量名的大括号

Using braces with dynamic variable names in PHP

提问by user1159454

I'm trying to use dynamic variable names (I'm not sure what they're actually called) But pretty much like this:

我正在尝试使用动态变量名称（我不确定它们的实际名称）但非常像这样：

for($i=0; $i<=2; $i++) {
    $("file" . $i) = file($filelist[$i]);
}

var_dump($file0);

The return is nullwhich tells me it's not working. I have no idea what the syntax or the technique I'm looking for is here, which makes it hard to research. $filelistis defined earlier on.

回报是null告诉我它不起作用。我不知道我在这里寻找的语法或技术是什么，这使得研究变得困难。$filelist之前定义过。

回答by Sarfraz

Wrap them in {}:

将它们包裹起来{}：

${"file" . $i} = file($filelist[$i]);

Working Example

工作示例

---

---

Using ${}is a way to create dynamic variables, simple example:

using${}是一种创建动态变量的方法，简单的例子：

${'a' . 'b'} = 'hello there';
echo $ab; // hello there

回答by John Slegers

Overview

概述

In PHP, you can just put an extra $in front of a variable to make it a dynamic variable :

在 PHP 中，您可以$在变量前添加一个额外的变量，使其成为动态变量：

$$variableName = $value;

While I wouldn't recommend it, you could even chain this behavior :

虽然我不推荐它，但您甚至可以将这种行为链接起来：

$$$$$$$$DoNotTryThisAtHomeKids = $value;

You can but are not forced to put $variableNamebetween {}:

您可以但不必被迫$variableName介于{}以下内容之间：

${$variableName} = $value;

Using {}is only mandatory when the name of your variable is itself a composition of multiple values, like this :

{}仅当变量的名称本身是多个值的组合时才强制使用，如下所示：

${$variableNamePart1 . $variableNamePart2} = $value;

It is nevertheless recommended to always use {}, because it's more readable.

尽管如此，还是建议始终使用{}，因为它更具可读性。

Differences between PHP5 and PHP7

PHP5和PHP7的区别

Another reason to always use {}, is that PHP5 and PHP7 have a slightly different way of dealing with dynamic variables, which results in a different outcome in some cases.

另一个始终使用{}, 的原因是 PHP5 和 PHP7 处理动态变量的方式略有不同，这在某些情况下会导致不同的结果。

In PHP7, dynamic variables, properties, and methods will now be evaluated strictly in left-to-right order, as opposed to the mix of special cases in PHP5. The examples below show how the order of evaluation has changed.

在 PHP7 中，动态变量、属性和方法现在将严格按照从左到右的顺序进行评估，而不是在 PHP5 中混合特殊情况。下面的示例显示了评估顺序是如何变化的。

Case 1 : $$foo['bar']['baz']

情况1 ： $$foo['bar']['baz']

• PHP5 interpetation : ${$foo['bar']['baz']}
• PHP7 interpetation : ${$foo}['bar']['baz']

• PHP5 解释： ${$foo['bar']['baz']}
• PHP7解释： ${$foo}['bar']['baz']

Case 2 : $foo->$bar['baz']

案例2： $foo->$bar['baz']

• PHP5 interpetation : $foo->{$bar['baz']}
• PHP7 interpetation : $foo->{$bar}['baz']

• PHP5 解释： $foo->{$bar['baz']}
• PHP7解释： $foo->{$bar}['baz']

Case 3 : $foo->$bar['baz']()

案例3： $foo->$bar['baz']()

• PHP5 interpetation : $foo->{$bar['baz']}()
• PHP7 interpetation : $foo->{$bar}['baz']()

• PHP5 解释： $foo->{$bar['baz']}()
• PHP7解释： $foo->{$bar}['baz']()

Case 4 : Foo::$bar['baz']()

案例4： Foo::$bar['baz']()

• PHP5 interpetation : Foo::{$bar['baz']}()
• PHP7 interpetation : Foo::{$bar}['baz']()

• PHP5 解释： Foo::{$bar['baz']}()
• PHP7解释： Foo::{$bar}['baz']()

回答by Joakim Johansson

Try using {}instead of ():

尝试使用{}代替()：

${"file".$i} = file($filelist[$i]);

回答by Tom

I do this quite often on results returned from a query..

我经常对查询返回的结果执行此操作。

e.g.

例如

// $MyQueryResult is an array of results from a query

foreach ($MyQueryResult as $key=>$value)
{
   ${$key}=$value;
}

Now I can just use $MyFieldname (which is easier in echo statements etc) rather than $MyQueryResult['MyFieldname']

现在我可以只使用 $MyFieldname （在 echo 语句等中更容易）而不是 $MyQueryResult['MyFieldname']

Yep, it's probably lazy, but I've never had any problems.

是的，它可能是懒惰的，但我从来没有遇到任何问题。

回答by corysus

Tom if you have existing array you can convert that array to object and use it like this:

Tom 如果您有现有数组，则可以将该数组转换为对象并像这样使用它：

$r = (object) $MyQueryResult;
echo $r->key;

回答by Murad

i have a solution for dynamically created variable value and combined all value in a variable.

我有一个动态创建的变量值的解决方案，并将所有值组合到一个变量中。

if($_SERVER['REQUEST_METHOD']=='POST'){
    $r=0;
    for($i=1; $i<=4; $i++){
        $a = $_POST['a'.$i];
        $r .= $a;
    }
    echo $r;
}

回答by M61Vulcan

I was in a position where I had 6 identical arrays and I needed to pick the right one depending on another variable and then assign values to it. In the case shown here $comp_cat was 'a' so I needed to pick my 'a' array ( I also of course had 'b' to 'f' arrays)

我所处的位置有 6 个相同的数组，我需要根据另一个变量选择正确的数组，然后为其赋值。在此处显示的情况下 $comp_cat 是 'a' 所以我需要选择我的 'a' 数组（我当然也有 'b' 到 'f' 数组）

Note that the values for the position of the variable in the array go after the closing brace.

请注意，数组中变量位置的值位于右大括号之后。

${'comp_cat_'.$comp_cat.'_arr'}[1][0] = "FRED BLOGGS";

${'comp_cat_'.$comp_cat.'_arr'}[1][1] = $file_tidy;

echo 'First array value is '.$comp_cat_a_arr[1][0].' and the second value is .$comp_cat_a_arr[1][1];

${'comp_cat_'.$comp_cat.'_arr'}[1][0] = "FRED BLOGGS";

${'comp_cat_'.$comp_cat.'_arr'}[1][1] = $file_tidy;

echo '第一个数组值是'.$comp_cat_a_arr[1][0]。' 第二个值是 .$comp_cat_a_arr[1][1];

回答by Vildan Bina

Try using {} instead of ():

尝试使用 {} 代替 ()：

${"file".$i} = file($filelist[$i]);