json对象的变换,新手求大神帮忙
程序员文章站
2022-05-22 17:58:15
...
json对象的转换,新手求大神帮忙
json格式如下
{"loginResult":{"code":"005","teamId":"1","password":"123","role":"员工","roleId":"2"}}
我用的CI框架 在models 里面定义了一个Member_Model对象,属性有code,teamId,password,role,roleId
怎样将上面的JSON数据转换成对象赋给Member_Model
------解决方案--------------------
------解决方案--------------------
------解决方案--------------------
如果是
$result = json_decode($buffer, true);
则
$member = $result['loginResult'];
如果是
$result = json_decode($buffer);
则
$member = $result->loginResult;
json格式如下
{"loginResult":{"code":"005","teamId":"1","password":"123","role":"员工","roleId":"2"}}
我用的CI框架 在models 里面定义了一个Member_Model对象,属性有code,teamId,password,role,roleId
怎样将上面的JSON数据转换成对象赋给Member_Model
------解决方案--------------------
$json = '{"loginResult":{"code":"005","teamId":"1","password":"123","role":"员工","roleId":"2"}}';
$arr = json_decode($json);
$mm = new Member_Model();
foreach ($arr->loginResult as $key => $value) {
$mm->$key = $value;
}
------解决方案--------------------
$s='{"loginResult":{"code":"005","teamId":"1","password":"123","role":"员工","roleId":"2"}}';
$o = json_decode($s);
print_r($o->loginResult);
------解决方案--------------------
如果是
$result = json_decode($buffer, true);
则
$member = $result['loginResult'];
如果是
$result = json_decode($buffer);
则
$member = $result->loginResult;
相关文章
相关视频