如何将这样的数组转换为对象?

[128] => Array
    (
        [status] => "Figure A.
 Facebook's horizontal scrollbars showing up on a 1024x768 screen resolution."
    )

[129] => Array
    (
        [status] => "The other day at work, I had some spare time"
    )

当前回答

我使用以下代码将Yaml文件关联数组解析为对象状态。

这将检查所有提供的数组中是否隐藏有对象,并将它们转换为对象。

    /**
     * Makes a config object from an array, making the first level keys properties a new object.
     * Property values are converted to camelCase and are not set if one already exists.
     * @param array $configArray Config array.
     * @param boolean $strict To return an empty object if $configArray is not an array
     * @return stdObject The config object
     */
    public function makeConfigFromArray($configArray = [],$strict = true)
    {
        $object = new stdClass();

        if (!is_array($configArray)) {
            if(!$strict && !is_null($configArray)) {
                return $configArray;
            }
            return $object;
        }

        foreach ($configArray as $name => $value) {
            $_name = camel_case($name);
            if(is_array($value)) {
                $makeobject = true;
                foreach($value as $key => $val) {
                    if(is_numeric(substr($key,0,1))) {
                        $makeobject = false;
                    }
                    if(is_array($val)) {
                        $value[$key] = $this->makeConfigFromArray($val,false);
                    }
                }
                if($makeobject) {
                    $object->{$name} = $object->{$_name} = $this->makeConfigFromArray($value,false);
                }
                else {
                    $object->{$name} = $object->{$_name} = $value;
                }

            }
            else {
                $object->{$name} = $object->{$_name} = $value;
            }
        }

        return $object;
    }

这将把yaml配置为

fields:
    abc:
        type: formfield
        something:
            - a
            - b
            - c
            - d:
                foo: 
                   bar

到由以下数组组成的数组:

array:1 [
  "fields" => array:1 [
    "abc" => array:2 [
      "type" => "formfield"
      "something" => array:4 [
        0 => "a"
        1 => "b"
        2 => "c"
        3 => array:1 [
          "d" => array:1 [
            "foo" => "bar"
          ]
        ]
      ]
    ]
  ]
]

以…为目标:

{#325
  +"fields": {#326
    +"abc": {#324
      +"type": "formfield"
      +"something": array:4 [
        0 => "a"
        1 => "b"
        2 => "c"
        3 => {#328
          +"d": {#327
            +"foo": "bar"
          }
        }
      ]
    }
  }
}

其他回答

快速攻击:

// assuming $var is a multidimensional array
$obj = json_decode (json_encode ($var), FALSE);

不漂亮,但很好用。

我肯定会用这样一种干净的方式:

<?php

class Person {

  private $name;
  private $age;
  private $sexe;

  function __construct ($payload)
  {
     if (is_array($payload))
          $this->from_array($payload);
  }


  public function from_array($array)
  {
     foreach(get_object_vars($this) as $attrName => $attrValue)
        $this->{$attrName} = $array[$attrName];
  }

  public function say_hi ()
  {
     print "hi my name is {$this->name}";
  }
}

print_r($_POST);
$mike = new Person($_POST);
$mike->say_hi();

?>

如果你提交:

你会得到这个:

我发现这更符合逻辑的比较以上的答案从对象应该用于他们的目的(封装可爱的小对象)。

此外,使用get_object_vars确保在被操纵的对象中没有创建额外的属性(您不希望汽车有姓,也不希望人有4个轮子)。

这需要PHP7,因为我选择使用lambda函数来锁定主函数中的'innerfunc'。lambda函数是递归调用的,因此需要:"use (&$innerfunc)"。你可以在PHP5中这样做,但不能隐藏innerfunc。

function convertArray2Object($defs) {
    $innerfunc = function ($a) use ( &$innerfunc ) {
       return (is_array($a)) ? (object) array_map($innerfunc, $a) : $a; 
    };
    return (object) array_map($innerfunc, $defs);
}

容易:

$object = json_decode(json_encode($array));

例子:

$array = array(
    'key' => array(
        'k' => 'value',
    ),
    'group' => array('a', 'b', 'c')
);

$object = json_decode(json_encode($array));

那么,以下是正确的:

$object->key->k === 'value';
$object->group === array('a', 'b', 'c')

我用了很简单的方法,

    $list_years         = array();
    $object             = new stdClass();

    $object->year_id   = 1 ;
    $object->year_name = 2001 ;
    $list_years[]       = $object;