Question

我一直在尝试为我的json数组找到正确的代码，该代码来自mysql数据库。所以要清楚，让我先发布代码。

json数组（来自下面的php代码）

{"query":"han","user_info":[{"username":"HannaShissler"},{"avatar":"\/\/en.wetube.org\/tubes\/uploads\/avatar\/no_avatar.png-50"},{"username":"Athanh"},{"avatar":"\/\/en.wetube.org\/tubes\/uploads\/avatar\/no_avatar.png-50"},{"username":"sabhan"},{"avatar":"\/\/en.wetube.org\/tubes\/uploads\/avatar\/no_avatar.png-50"}]}

PHP

$search_sql = $db->query("SELECT * FROM users WHERE username like '%$query%' order by id DESC LIMIT 5");        
$data = array();
while($array = $db->fetch_assoc($search_sql)){
    $data[]['username'] = $array['username'];
    $data[]['avatar'] = DOMAIN.TUBES.UPLOADS.AVATAR.resizer($array['avatar'],'50','');
}
$data = array(
'query'=>$query,
'user_info'=>$data
);

echo json_encode($data);

JS / AJAX / JSON

success: function(data){

$.each(data.user_info, function(index,value) {
    alert(value.username);
  }); 

}

现在我想要获得的是user_info子数组的用户名。如果有另一种重写PHP代码的方法，因为我觉得我做错了。请帮助这些人，我依赖你。或者我将不得不回到html而不是json数据:(

Answer 1

我会改变

{"username":"HannaShissler"},{"avatar":"\/\/en.wetube.org\/tubes\/uploads\/avatar\/no_avatar.png-50"}

到

{"username":"HannaShissler","avatar":"\/\/en.wetube.org\/tubes\/uploads\/avatar\/no_avatar.png-50"}

（每个用户一个对象，而不是将头像和用户名放在不同的对象中）

执行此操作的代码将是

$search_sql = $db->query("SELECT * FROM users WHERE username like '%$query%' order by id DESC LIMIT 5");        
$data = array();
while($array = $db->fetch_assoc($search_sql)){
    $data[]=array(
        'username' => $array['username'];
        'avatar' => DOMAIN.TUBES.UPLOADS.AVATAR.resizer($array['avatar'],'50','')
    );
}
$data = array(
'query'=>$query,
'user_info'=>$data
);

echo json_encode($data);

如何通过键名检索json子数组值

1 个答案: