【发布时间】:2019-11-08 04:03:48
【问题描述】:
我刚学过php
我有一个单独的数据库表,我想将该表与只有 1 个参数的 JSON 输出结合起来,但结果输出是错误的,如何将我的 JSON 更改为正确且易于使用的 json。
谢谢
json 输出:
[
[
{
"id_service": "3",
"reference_number": "",
"tracking_number": "RJC-0000-0001",
"kd_inbound": "INB-1000-0001",
"tgl_inbound": "2019-11-07 00:00:00",
"status_inb": "1"
}
],
[
{
"id_service": "3",
"reference_number": "",
"kd_outbag": "BAG-1468-0002",
"tanggal_outbag": "2019-11-07 00:00:00",
"status_outbag": "1"
}
],
[
{
"id_service": "3",
"reference_number": "",
"kd_outbound": "OTB-1826-0001",
"tgl_outbound": "2019-11-07 17:04:49",
"status_otb": "1"
}
],
]
这是我的代码
public function awb_get() {
$id = $this->get('tracking_number');
$res= array(
$this->M_tarif->tampil_status_inbound($id),
$this->M_tarif->tampil_status_otboundbag($id),
$this->M_tarif->tampil_status_otboundori($id),
$this->M_tarif->tampil_status_indes($id),
$this->M_tarif->tampil_status_outdes($id),
$this->M_tarif->tampil_status_runsheet($id),
$this->M_tarif->tampil_db_service_status($id)
);
$this->response($res, 200);
}
我想要这样
{
"status": 200,
"error": false,
"awb": [
{
"tracking_number": "RJC-0000-0004",
"status": "order",
"tanggal": "2019-10-30"
},
{
"tracking_number": "RJC-0000-0004",
"status": "Inbound to origin",
"tanggal": "2019-11-03"
}
]
}
【问题讨论】:
-
您必须使用 json_decode() 将其转换为数组,而不是使用 forloop
-
谢谢,能举个例子吗?
-
你的问题有点不清楚,你要结合什么表?
-
您的 JSON 输出无效Jsonlint
-
有没有办法让json输出正确?
标签: php json codeigniter