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

使用我创建的这个函数:

function buildObject($class,$data){
    $object = new $class;
    foreach($data as $key=>$value){
        if(property_exists($class,$key)){
            $object->{'set'.ucfirst($key)}($value);
        }
    }
    return $object;
}

用法:

$myObject = buildObject('MyClassName',$myArray);

其他回答

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"
          }
        }
      ]
    }
  }
}

多维数组转换为对象。此代码用于转换必应搜索API的尝试和捕获方法。

try {
        // Perform the Web request and get the JSON response
        $context = stream_context_create($options);
        $results = file_get_contents($url . "?cc=" . $country . "&category=" . $type, false, $context);
        $results = json_decode($results);
        return response()->json($results);
    } catch (\Exception $e) {
        $results = array('value' => array(
                (object) array(
                    "name" => "Unable to Retrive News",
                    "url" => "http://www.sample.com/",
                    "image" => (object) array("thumbnail" => (object) array("contentUrl" => "")),
                    "publishedAt" => "",
                    "description" => "")
            )
        );
        $results = (object) $results;
        return response()->json($results);
    }

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. ..."

我也有这个问题,但我注意到json_decode将JSON数组转换为对象。

所以,我通过使用json_encode($PHPArray)来实现我的解决方案,它返回对象的JSON字符串,然后我用Json_decode($string)解码字符串,它将返回一个完美的结构化对象。 速记

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

Or

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

这里有三种方法:

Fake a real object: class convert { public $varible; public function __construct($array) { $this = $array; } public static function toObject($array) { $array = new convert($array); return $array; } } Convert the array into an object by casting it to an object: $array = array( // ... ); $object = (object) $array; Manually convert the array into an object: $object = object; foreach ($arr as $key => $value) { $object->{$key} = $value; }