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

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);
}

它的方法很简单,这将为递归数组创建一个对象:

$object = json_decode(json_encode((object) $yourArray), FALSE);

根据你需要的位置和访问对象的方式有不同的方法。

例如:只需对它进行类型转换

$object =  (object) $yourArray;

然而,最兼容的方法是使用一个实用程序方法(还不是PHP的一部分),它实现了基于指定类型的字符串的标准PHP强制转换(或者忽略它,只是去引用值):

/**
 * dereference a value and optionally setting its type
 *
 * @param mixed $mixed
 * @param null  $type (optional)
 *
 * @return mixed $mixed set as $type
 */
function rettype($mixed, $type = NULL) {
    $type === NULL || settype($mixed, $type);
    return $mixed;
}

您案例中的使用示例(在线演示):

$yourArray = Array('status' => 'Figure A. ...');

echo rettype($yourArray, 'object')->status; // prints "Figure A. ..."

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;

您可以简单地使用类型强制转换将数组转换为对象。

// *convert array to object* Array([id]=> 321313[username]=>shahbaz)
$object = (object) $array_name;

//now it is converted to object and you can access it.
echo $object->username;