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"
)
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;