浅析php中json_encode()和json_decode()_PHP教程
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2022-05-17 21:31:42
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json_encode()
$arr = array ('a'=>'a','b'=>'b','c'='c','d'=>'d','e'='e');
echo json_encode($arr);
class person
{
public $name;
public $age;
public $height;
function __construct($name,$age,$height)
{
$this->name = $name;
$this->age = $age;
$this->height = $height;
}
}
$json = '{"a":"hello","b":"world","c":"zhangsan","d":20,"e":170}';
var_dump(json_decode($json));
$json = '{"a":"hello","b":"world","c":"zhangsan","d":20,"e":170}';
var_dump(json_decode($json,ture));
该函数主要用来将数组和对象,转换为json格式。
复制代码 代码如下:
$arr = array ('a'=>'a','b'=>'b','c'='c','d'=>'d','e'='e');
echo json_encode($arr);
输出结果:
json只接受utf-8编码的字符,json_encode()的参数必须是utf-8编码。
复制代码 代码如下:
class person
{
public $name;
public $age;
public $height;
function __construct($name,$age,$height)
{
$this->name = $name;
$this->age = $age;
$this->height = $height;
}
}
$obj = new person("zhangsan",20,100);
$foo_json = json_encode($obj);
echo $foo_json;
输出结果:
当类中的属性为私有变量的时候,则不会输出。
json_decode()
该函数用于将json文本转换为相应的PHP数据结构。
复制代码 代码如下:
$json = '{"a":"hello","b":"world","c":"zhangsan","d":20,"e":170}';
var_dump(json_decode($json));
输出结果:
通常情况下,json_decode()总是返回一个PHP对象。
转成数组的:
复制代码 代码如下:
$json = '{"a":"hello","b":"world","c":"zhangsan","d":20,"e":170}';
var_dump(json_decode($json,ture));