Question

我需要一些关于PHP json的帮助。关于这件事我还是新手。我尝试了其他方法，但似乎没有成功。我有这种json格式：

{
        "schedule_backup": [
            {
                "class_time_id_fk": "1",
                "student_id_number": "AR0001",
                "teacher_id_number": "ACAD-0091",
                "class_type_id_fk": "1",
                "room_assignment_id_fk": "4",
                "books_materials_id_fk": "225",
                "class_level_id_fk": "23",
                "subject_id_fk": "57"
            },
            {
                "class_time_id_fk": "2",
                "student_id_number": "AR0001",
                "teacher_id_number": "ACAD-0049",
                "class_type_id_fk": "1",
                "room_assignment_id_fk": "41",
                "books_materials_id_fk": "211",
                "class_level_id_fk": "4",
                "subject_id_fk": "58"
            }
       ]
    }

我想在桌子上展示它们。这是我的代码：

我的观点

echo $jsonString = $this->MClassSchedule->parse_schedule_backup();

$jsonArray = json_decode($jsonString,true);

echo $jsonArray['schedule_backup']['class_time_id_fk'];

模型

function parse_schedule_backup(){
    $this->db->select('*');
    $this->db->from('class_schedule_backup');
    $query = $this->db->get();

    foreach ($query->result() as $rows) {

        return $rows->data;

    }

}

Answer 1

除非你以异步方式或通过API传递它，否则将它转换为JSON然后返回它似乎很麻烦......只需像下面的模型一样返回数据数组。

<强>模型

function parse_schedule_backup(){
    $this->db->select('*');
    $this->db->from('class_schedule_backup');
    $query = $this->db->get();

    // Return associative data array
    return $query->result_array();
}

然后在视图中，只循环遍历表结构内的行。人们对模板表有不同的风格，但下面有一种方法（显然不包括所有列）。

我的观点

<?php
// You probably want to do this in the controller and pass the data to the view
$data = $this->MClassSchedule->parse_schedule_backup();
// Create table header
?>
<table>
   <thead>
       <tr>
           <th>Class ID</th>
           <th>Teacher ID</th>
           <th>Student</th>
       </tr>
   </thead>
   <tbody>
<?php
    // You probably want to check for a non-empty array and provide an error if needed
    foreach($data as $row){
        $json = json_decode($row['data'], TRUE);
?>
        <tr>
            <td><?= $json['schedule_backup']['timeID_1']['class_time_id_fk'] ?></td>
            <td><?= $json['schedule_backup']['timeID_1']['student_id_number'] ?></td>
            <td><?= $json['schedule_backup']['timeID_1']['teacher_id_number'] ?></td>
        </tr>
<?php 
    }
?>
   </tbody>
</table>

如何使用codeigniter显示json的数据

1 个答案: