将json转换为对象php

我刚学过php

我有一个单独的数据库表，我想将该表与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"
    }
 ],
]

这是我的密码

 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);
}

我想这样进去

{
"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"
    }
]
}

---

您的json响应不正确。 因此，在发送到函数之前，首先必须重置JSON。我制作了一个示例来重新排列JSON并在stdClass对象中转换它。请检查下面的代码以转换您的响应

$arr ='[
 [
    {
        "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"
    }
 ],
]';

$result = str_replace(array('[',']','\n'), '',htmlspecialchars(json_encode($arr), ENT_NOQUOTES));
$str = preg_replace('/\\\"/',"\"", $result);

$json = '[';
$json .= substr($str, 2,-1); // Substring -1 character from the end of the json variable, this will be the trailing comma. 
$json .= ']';
$jsonData = preg_replace("/,(?!.*,)/", "", $json);

echo "<pre>";
print_r(json_decode($jsonData));
echo "</pre>";

---

你必须使用FORLOOP将其转换为数组谢谢，你能举个例子吗？你的问题有点不清楚，你想合并哪个表？你的JSON输出无效，有没有解决方案来生成正确的JSON输出？在这里，我只是用正确的格式将JSON响应转换为正确的格式。使用与演示json数组相同的代码。在$arr数组变量中有什么？与您在json响应中提到的相同？

Array
(
    [0] => stdClass Object
        (
            [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
        )

    [1] => stdClass Object
        (
            [id_service] => 3
            [reference_number] => 
            [kd_outbag] => BAG-1468-0002
            [tanggal_outbag] => 2019-11-07 00:00:00
            [status_outbag] => 1
        )

    [2] => stdClass Object
        (
            [id_service] => 3
            [reference_number] => 
            [kd_outbound] => OTB-1826-0001
            [tgl_outbound] => 2019-11-07 17:04:49
            [status_otb] => 1
        )

)