Question

我想从我的查询中生成json数据。但我不能像我预期的那样得到我的json数据。这是我试过的代码。

<?php 
        $this->db->select('campus.*, class.*');
        $this->db->from('campus');
        $this->db->join('class', 'campus.id = class.campus_id');
        $campus_info = $this->db->get()->result_array();
        foreach ($campus_info as $row) {
            $var[]['id'] = $row['id'];
            $var[]['campus_name'] = $row['campus_name'];
            $var[]['class_id'] = $row['class_id'];          
            $var[]['class_name'] = $row['name'];
        }
        $json_data = json_encode($var, JSON_PRETTY_PRINT);
        echo '<pre>';
        echo $json_data; 
        echo '</pre>';
        exit();

我的查询结果是

Array
(
    [0] => Array
        (
            [id] => 4
            [campus_name] => Test Campus Name
            [institute_name] => Test Institute Name 1
            [class_id] => 11
            [name] => STD - VIII
            [campus_id] => 4
            [name_numeric] => 8
            [teacher_id] => 6
        )

    [1] => Array
        (
            [id] => 4
            [campus_name] => Test Campus Name
            [institute_name] => Test Institute Name 1
            [class_id] => 15
            [name] => A' Level
            [campus_id] => 4
            [name_numeric] => 12
            [teacher_id] => 6
        )

想让它像json一样

[{
    "id": "7",
    "campus_name": "Azimpur",
    "class_id": "9",
    "class_name": "STD - VI"
}
{
    "id": "8",
    "campus_name": "Azimpur",
    "class_id": "10",
    "class_name": "STD - VII"
}]

但是得到

[
    {
        "id": "4"
    },
    {
        "campus_name": "Test Campus Name"
    },
    {
        "class_id": "11"
    },
    {
        "class_name": "STD - VIII"
    },
    {
        "id": "4"
    },
    {
        "campus_name": "Test Campus Name"
    },

Answer 1

这是你的问题：

foreach ($campus_info as $row) {
  $var[]['id'] = $row['id'];
  $var[]['campus_name'] = $row['campus_name'];
  $var[]['class_id'] = $row['class_id'];          
  $var[]['class_name'] = $row['name'];
}

在这里，您要在$var变量的最高级别添加4个元素。

你应该只添加一个元素：

foreach ($campus_info as $row) {
  $var[] = [
    'id' => $row['id'],
    'campus_name' => $row['campus_name'],
    'class_id' => $row['class_id'],     
    'class_name' => $row['name']
  ];
}

如果您只选择所需的列，则可以将其缩小为：

foreach ($campus_info as $row) {
  $var[] = $row;
}

如果您的数据库API具有fetchAll方法，则可能更进一步。

从活动记录中获取json数据

1 个答案: