在PHP多维数组中返回引用并操作特定变量。

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英文:

Returning reference and manipulating specific variable in PHP multidimensional array

问题

我正在尝试从多维全局数组的特定位置返回引用,然后进行操作... 这是一个深度嵌套的数组。

我试过各种方法:

$foo = &get('bar');

$foo['something'] = 'for nothing';
 
var_dump($array); // 未受影响

function &get($search) {
    global $array;
    foreach(...) {
        foreach(...) {
            foreach(... as $key => &$val) {
               if($val === $search) return $val;
            }
        }
    }
}

这在PHP中可能不可行,但希望我错了,只是错过了一些明显的东西。

如果有更容易存储和操作复杂对象的方法,我并不一定要使用多维数组,只需要尽可能快。

谢谢任何建议!

英文:

I'm trying to return a reference from a specific position in a multidimensional global array and then manipulate it ... it's a deep nested array.

I'm trying to avoid having to iterate the array in every function that needs to manipulate a specific part.

I've tried all matter of this:

$foo = &get('bar');

$foo['something'] = 'for nothing';
 
var_dump($array); // unaffected

function &get($search) {
    global $array;
    foreach(...) {
        foreach(...) {
            foreach(... as $key => &val) {
               if($val === $search) return &val;
            }
        }
    }
}

This may not even be possible in PHP, but hoping I'm wrong and just missing something obvious.

Also, not married to a multidimensional array if there is an easier way to store and manipulate complex objects, it just needs to be as fast as possible.

Thanks for any advice!

答案1

得分: 1

I've fixed the syntax errors.

你的语法错误已经修复。

You need to use references for the iteration variables in all the foreach loops.

在所有的 foreach 循环中,你需要使用引用来处理迭代变量。

You don't need to use & when you're returning $val, just return $val;.

当你返回 $val 时,不需要使用 &,只需写成 return $val;

To update, you should just assign directly to $foo, since it doesn't contain an array.

要进行更新,你只需要直接赋值给 $foo,因为它不包含数组。

英文:

I've fixed the syntax errors.

You need to use references for the iteration variables in all the foreach loops.

You don't need to use & when you're returning $val, just return $val;.

To update, you should just assign directly to $foo, since it doesn't contain an array.

<?php
$array = [[["foo", "bar"]]];
$foo = &get('bar');

$foo = 'for nothing';

var_dump($array);

function &get($search) {
    global $array;
    foreach($array as &$level1) {
        foreach($level1 as &$level2) {
            foreach($level2 as &$val) {
               if($val === $search) return $val;
            }
        }
    }
}

Output:

array(1) {
  [0]=>
  array(1) {
    [0]=>
    array(2) {
      [0]=>
      string(3) "foo"
      [1]=>
      &string(11) "for nothing"
    }
  }
}

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  • 本文由 发表于 2023年8月5日 05:07:56
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