我正在尝试使用PHP / PDO从MySQL提取数据并将其格式化为嵌套的JSON数组。
这是我得到的:
{
"host1": [
{
"vmnic_name": "vmnic0",
"switch_name": "switch1",
"port_id": "GigabitEthernet1\/0\/1"
},
{
"vmnic_name": "vmnic1",
"switch_name": "switch1",
"port_id": "GigabitEthernet1\/0\/2"
}
],
"host2": {
"2": {
"vmnic_name": "vmnic0",
"switch_name": "switch1",
"port_id": "GigabitEthernet1\/0\/3"
},
"3": {
"vmnic_name": "vmnic1",
"switch_name": "switch1",
"port_id": "GigabitEthernet1\/0\/4"
}
}
}
我想说“ host_name”:“ host1”等,而不只是“ host1”。对于第一个主机之后的主机,不要像第一个主机那样具有“ 2”或“ 3”之类的数字。
这是我的代码:
$arr = array();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
foreach ($result as $key => $item) {
$arr[$item['host_name']][$key] = array(
'vmnic_name'=>$item['vmnic_name'],
'switch_name'=>$item['switch_name'],
'port_id'=>$item['port_id']
);
}
echo json_encode($arr, JSON_PRETTY_PRINT);
任何帮助将不胜感激。
如果只选择查询中的那些列,那么就这么简单:
foreach ($result as $item) {
$arr[$item['host_name']][] = $item;
}
如果出于任何原因必须选择更多列以供以后使用,则只需插入host_name
并删除$key
作为索引:
foreach ($result as $key => $item) {
$arr[$item['host_name']][] = array(
'host_name'=>$item['host_name'],
'vmnic_name'=>$item['vmnic_name'],
'switch_name'=>$item['switch_name'],
'port_id'=>$item['port_id']
);
}
这很容易。将其解码为一个数组,然后将其粘贴到另一个数组中。
$array = json_decode($json, true);
$x = [];
$x['host_name'] = $array;
var_dump(json_encode($x, JSON_PRETTY_PRINT));
哪个会给你:
string(747) "{ "host_name": { "host1": [ { "vmnic_name": "vmnic0", "switch_name": "switch1", "port_id": "GigabitEthernet1\/0\/1" }, { "vmnic_name": "vmnic1", "switch_name": "switch1", "port_id": "GigabitEthernet1\/0\/2" } ], "host2": { "2": { "vmnic_name": "vmnic0", "switch_name": "switch1", "port_id": "GigabitEthernet1\/0\/3" }, "3": { "vmnic_name": "vmnic1", "switch_name": "switch1", "port_id": "GigabitEthernet1\/0\/4" } } } }"