Php 如何从结果集生成嵌套数组结构?
我有一个Json数组事件Php 如何从结果集生成嵌套数组结构?,php,nested,resultset,Php,Nested,Resultset,我有一个Json数组事件 [ { "id": "4", "event_name": "Harliquins 7s", "event_description": "Ruggby game", "event_date": null, "event_venue": "UFA grounds", "event_company": "Harliquins", "event_image":
[
{
"id": "4",
"event_name": "Harliquins 7s",
"event_description": "Ruggby game",
"event_date": null,
"event_venue": "UFA grounds",
"event_company": "Harliquins",
"event_image": "http://www.aal-europe.eu/wp-content/uploads/2013/12/events_medium.jpg",
"event_ticket_no": "200",
"paybill": "25666",
"status": "0"
},
{
"id": "5",
"event_name": "christie &s",
"event_description": "Ruggby",
"event_date": "1-2-2917",
"event_venue": "KISUMU ground",
"event_company": "Kenya Games",
"event_image": "N/A",
"event_ticket_no": "400",
"paybill": "79000",
"status": "0"
}
]
我也有这个选择
[
{
"id": "4",
"event_id": "5",
"options_id": "1",
"seasonal": "1",
"amount": "300"
},
{
"id": "5",
"event_id": "5",
"options_id": "2",
"seasonal": "1",
"amount": "400"
}
]
我想得到这个结果
[
{
"id": "4",
"event_name": "Harliquins 7s",
"event_description": "Ruggby game",
"event_date": null,
"event_venue": "UFA grounds",
"event_company": "Harliquins",
"event_image": "http://www.aal-europe.eu/wp-content/uploads/2013/12/events_medium.jpg",
"event_ticket_no": "200",
"paybill": "25666",
"status": "0"
},
{
"id": "5",
"event_name": "christie &s",
"event_description": "Ruggby",
"event_date": "1-2-2917",
"event_venue": "KISUMU ground",
"event_company": "Kenya Games",
"event_image": "N/A",
"event_ticket_no": "400",
"paybill": "79000",
"status": "0",
"Options:" [
{
"id": "4",
"event_id": "5",
"options_id": "1",
"seasonal": "1",
"amount": "300"
},
{
"id": "5",
"event_id": "5",
"options_id": "2",
"seasonal": "1",
"amount": "400"
}
]
}
]
这是我的密码:
while( $row = $result->fetch_assoc()){
$new_array[] =$row;
$new_array['options'] =getTicketOptions($row['id']);
}
echo json_encode($new_array);
每个事件对象都有一个选项数组。您已经差不多做到了。您只需立即分配$row和选项:
while( $row = $result->fetch_assoc()){
$new_array[] = array($row, "Options" => getTicketOptions($row['id']));
}
echo json_encode($new_array);
原因是使用
$new\u array[]
可以自动设置一个新键,而不知道其值。因此,您以后不能简单地将某些内容添加到此记录中。同时执行这两项操作可以为您解决问题。在我看来,您只需要合并两个对象:
$mergedObject = (object) array_merge((array) $events, (array) $options);
我想您想在最后一个位置添加第二个数组??如果是,则按照回答 第一阵列
[{"id":"4","event_name":"Harliquins 7s"},{"id":"5","event_name":"Harliquins 7s"}]
第二阵列
[{"id":"4","event_id":"5","options_id":"1"},{"id":"4","event_id":"5","options_id":"1"}]
必需数组
[{"id":"4","event_name":"Harliquins 7s"},{"id":"5","event_name":"Harliquins 7s","options":[{"id":"4","event_id":"5","options_id":"1"},{"id":"4","event_id":"5","options_id":"1"}]}]
Php代码
$last_key = end(array_keys($data));
foreach ($data as $key => $value) {
if($key == $last_key){
$data[$key]['options'] = $data2;
}
}
echo json_encode($data);
我使用和函数得到相同的结果,您可以在此处尝试我的代码:,它始终在事件数组中创建选项属性:
$events = json_decode($events, true);
$options = json_decode($options, true);
$result = array_map( function ($item) use ($options) {
$item['options'] = array_filter($options, function ($option) use ($item) {
return $item['id'] == $option['event_id'];
});
return $item;
}, $events);
print_r($result);
我担心您的while循环的效率可能会重复。
getTicketOptions()
在做什么?它是在调用数据库查询吗?或者它正在访问一个静态数组?还是别的什么?如果这是进行迭代查询,那么单连接查询可能是更好的方法。请澄清您代码的这一方面。您不打算从找到的重复页面复制答案。这会在SO上产生冗余信息。哦,好的。那是不是意味着我不该费心回答?或者我应该把它作为一个评论放进去吗?只需要标记/投票以重复方式结束。这有助于OP找到正确的解决方案。减少浪费的志愿者时间(阅读问题并回答已回答的问题)。OP应在提问前进行搜索和研究——标记为重复表示未进行详尽的研究(由于缺乏知识或懒惰)。这一主题是不同用户之间的争论点;如果你想从更广阔的角度来看待这件事,我建议你搜索Meta。你会看到更多的理由和一些反驳。