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
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;