这是我的数据库:
week, subbizname, devicetype
20141203, common, PC
20141203, unknown, PC
20141210, KRsinKD, SP
20141210, unknown, PC
20141217, Unknown, SP
20141217, Chintai, TAB
....
我正在尝试获取每个独特的夫妇devicetype /周的记录数。 例如:
array(
20141203 => array(
'PC'=>2,
'TAB'=>0,
'SP'=>0
),
20141210 => array(
'PC'=>1,
'TAB'=>0,
'SP'=>1
),
...
.....
)
更新:
我使用了查询:
$data = $this->Test->find('all', array(
'conditions'=>array(
'OR'=>array(
array('devicetype'=>'PC'),
array('devicetype'=>'SP'),
array('devicetype'=>'TAB'),
)
),
'fields'=>"devicetype,week,COUNT(devicetype) AS countDevice",
'group'=>"week,devicetype"
));
事情是,它返回这样的东西:
(int) 0 => array(
'Test' => array(
'devicetype' => 'PC',
'week' => '20141126'
),
(int) 0 => array(
'countDevice' => '34844'
)
),
(int) 1 => array(
'Test' => array(
'devicetype' => 'SP',
'week' => '20141126'
),
(int) 0 => array(
'countDevice' => '32401'
)
),
(int) 2 => array(
'Test' => array(
'devicetype' => 'TAB',
'week' => '20141126'
),
(int) 0 => array(
'countDevice' => '4256'
)
),
(int) 3 => array(
'Test' => array(
'devicetype' => 'PC',
'week' => '20141203'
),
(int) 0 => array(
'countDevice' => '96564'
)
),
(int) 4 => array(
'Test' => array(
'devicetype' => 'SP',
'week' => '20141203'
),
(int) 0 => array(
'countDevice' => '97450'
)
),
但我无法获得预期的结果。 当然必须有更好的方法。 我该如何解决这个问题?
答案 0 :(得分:0)
使用$ data变量,您可以:
$d = array();
foreach ($data as $value) {
$week = $value['Test']['week'];
if (isset($d[$week])) {
$d[$week][$value['Test']['devicetype']] = $value[0]['countDevice'];
} else {
$d[$week] = array($value['Test']['devicetype'] => $value[0]['countDevice']);
}
}
print_r($d); // $d will contain the data in your required format