Question

我有一张桌子，我希望从数据库中获取数据并将其返回到 JSON fromat 日期老虎钳，下面是我的表格：

id  userId      Date      Time   record
1     1      15-Oct-2017  3:50    152
2     1      15-Oct-2017  4:30    142
3     1      16-Oct-2017  8:50    130
4     2      15-Oct-2017  2:00    90
5     2      15-Oct-2017  4:50    154
6     2      15-Oct-2017  5:00    120

我创建了一个函数，在其中我从数据库调用数据并返回 JSON fromat

中的输出

public function getRecord()
    {
        $userId = $this->input->get('userId');
        $this->db->from('xyz');
        $this->db->where('userId',$userId);
        $record = $this->db->get()->result();

        return $this->output->set_output(json_encode(array(     
                'status' => 'Ok',
                'statusCode' =>'801',
                'response' => $record

            )));
    }

它给我这样的东西，这是我不需要的（我知道我没有以正确的方式去做）

{
  "status": "Ok",
  "statusCode": "801",
  "response": Array[3][
    {
      "id": "1",
      "userId": "1",
      "date": "15-Oct-2017",
      "time": "3:50",
      "record": "152"
    },
    {
      "id": "2",
      "userId": "1",
      "date": "15-Oct-2017",
      "time": "4:30",
      "record": "142"
    },
    {
      "id": "3",
      "userId": "1",
      "date": "16-Oct-2017",
      "time": "8:50",
      "record": "130"
    }
  ]
}

但是我想要这样的东西，输出将被放入日期虎钳

{
"status": "Ok",
"statusCode": "801",
"response": Array[3][

[
"date": "15-Oct-2017"
{ 
"id": "1",
"userId": "1",
"date": "15-Oct-2017",
"time": "3:50",
"record": "152"
},
{  
"id": "2",
"userId": "1",
"date": "15-Oct-2017",
"time": "4:30",
"record": "142"
}
],

[
"date": "16-Oct-2017"
{
"id": "3",
"userId": "1",
"date": "16-Oct-2017",
"time": "8:50",
"record": "130"
}
]

]
}

Answer 1

虽然您提供的结果json似乎没有效果，因此，我认为您希望按日期分组的记录，尝试循环记录数组，按日期将它们分组到另一个数组上：

<?php

    public function getRecord()
    {
        $userId = $this->input->get('userId');
        $this->db->from('xyz');
        $this->db->where('userId',$userId);
        $record = $this->db->get()->result();
        $orderedRecord = [];

        foreach ($record as $r) {
            if (!isset($orderedRecord[$r->Date])) {
                $orderedRecord[$r->Date] = [];
            }

            $orderedRecord[$r->Date][] = $r;
        }

        return $this->output->set_output(json_encode(array(     
            'status' => 'Ok',
            'statusCode' =>'801',
            'response' => $orderedRecord
        )));
    }
?>

如何从Codeigniter中的Date获取数据库组中的JSON数据

1 个答案: