将MySQLI fetch_assoc()压缩为每列一个数组 - Compact a MySQLI fetch_assoc() to one array per column

Compact a MySQLI fetch_assoc() to one array per column

感谢检测数据类型，同时使用fetch_array与MySQLi和输出属性与PHP5和方法链，我能够使用MySQLi和方法链在MySQL查询中填充json对象。

  /// Populate an array with the results;
  public function populate(){
    ($rows=array());
    while($row=$this->result->fetch_assoc())
      $rows[]=$row;
    $this->rows=$rows;
    $this->last=$this->rows;
    return $this;
  }

我获得

[
  {
      "id": 1,
      "datein": "2012-06-06 09:59:05"
  },
  {
      "id": 2,
      "datein": "2012-06-06 11:32:45"
  },
  {
      "id": 3,
      "datein": "2012-06-07 00:47:19"
  }
]

如何获得

{
  "id": [1,2,3]
  "datein": ["2012-06-06 09:59:05","2012-06-06 11:32:45","2012-06-07 00:47:19"]
}

，以便有一个替代的和紧凑的版本的结果?

非常感谢你的帮助!

编辑:

感谢你的帮助，我准备了这种mysql包装和两个抓取方法提供:http://ross.iasfbo.inaf.it/gloria/decibel-class

在数组中初始化2个数组，一个用于保存id，另一个用于保存datein

public function populate(){
    $rows = array('id'=>array(), 'datein'=>array());
    while($row=$this->result->fetch_assoc())
      $rows['id'][] = $row['id'];
      $rows['datein'][] = $row['datein'];
    $this->rows = $rows;
    $this->last = $this->rows;
    return $this;
}

您可以将代码更改为:

/// Populate an array with the results;
public function populate() {
    $rows = array();
    $this->result = array();
    while ($row = $this->result->fetch_assoc())
        foreach ($row as $field => $value)
            $this->result[$field][] = $value;
    return $this;
}