我从带有两个联接的MySQL查询中得到以下结果。
Array (
[0] => Array ( [place_id] => 1 [place] => Berlin [lat] => 52.519 [lon] => 13.406 [id] => 1 [pname] => Firstschool [typ] => 0 [s_id] => 32 [fac] => history)
[1] => Array ( [place_id] => 1 [place] => Berlin [lat] => 52.519 [lon] => 13.406 [id] => 1 [pname] => Secondschool [typ] => 0 [s_id] => 33 [fac] => math)
[2] => Array ( [place_id] => 1 [place] => Berlin [lat] => 52.519 [lon] => 13.406 [id] => 1 [pname] => Secondschool [typ] => 0 [s_id] => 33 [fac] => english)
)
数据在某些方面是多余的，我需要这样做：
Array (
[Berlin] => Array ( [lat] => 52.519
[lon] => 13.406
[schools] => Array([0]=> Firstschool [1]=>Secondschool)
)
[OtherCity] => Array ( ... )
)
首先，这是可以的还是存在更好的解决方案？=）第二如何将其拆分以获得所需的结果。
我在下面的代码片段中尝试过它，但它并没有按预期工作。
foreach($viewmodel as $item) {
$data[$item['place']][] = $item['pname'];
}
结果是：
Array ( [Berlin] => Array ( [0] => Firstschool [1] => Firstschool [2] => Firstschool ))
没那么有用。）
我希望我所需要的是可以理解的。也许有人对如何解决这个问题有个好主意。
谢谢你抽出时间。
我认为你走的是正确的道路，只需要填写更多的细节：
$cities = Array (
Array ( 'place_id' => 1, 'place' => 'Berlin', 'lat' => 52.519, 'lon' => 13.406, 'id' => 1, 'pname' => 'Firstschool', 'typ' => 0, 's_id' => 32, 'fac' => 'history'),
Array ( 'place_id' => 1, 'place' => 'Berlin', 'lat' => 52.519, 'lon' => 13.406, 'id' => 1, 'pname' => 'Secondschool', 'typ' => 0, 's_id' => 33, 'fac' => 'math'),
Array ( 'place_id' => 1, 'place' => 'Berlin', 'lat' => 52.519, 'lon' => 13.406, 'id' => 1, 'pname' => 'Secondschool', 'typ' => 0, 's_id' => 33, 'fac' => 'english'),
);
// gather the transformed array in a new array
$out = array();
foreach ($cities as $city) {
// the first time we see the place
if (!isset($out[$city['place']])) {
// copy over what you want to keep
$out[$city['place']] = array(
'lat' => $city['lat'],
'lon' => $city['lon'],
'schools' => array($city['pname']),
);
} // only add $city['pname'] if we don't have it already
elseif (!in_array($city['pname'], $out[$city['place']]['schools'])) {
// we already seen this place, just add to the schools
$out[$city['place']]['schools'][] = $city['pname'];
}
}
对于聚集教师的问题，使用学校名称作为顶级数组的"学校"键中数组的键，按如下方式填充：（仍跳过重复项）：
foreach ($a as $city) {
if (!isset($out[$city['place']])) {
$out[$city['place']] = array(
'lat' => $city['lat'],
'lon' => $city['lon'],
'schools' => array($city['pname'] => array($city['fac'])),
);
} else {
// for convenience and readability, introducing some variables
$schools = &$out[$city['place']]['schools'];
$pname = $city['pname'];
$fac = $city['fac'];
// if we didn't see this school yet, add it with it's faculty
if (!isset($schools[$pname])) {
$schools[$pname] = array($fac);
} // if we did see this school before but the faculty is new, add it under the school's key
else if (!in_array($fac, $schools[$pname])) {
$schools[$pname][] = $fac;
}
}
}
如果你是对的，你必须以某种方式迭代数组。从我看到的数组中，假设所有学校的纬度和经度都相同，覆盖不会有什么影响，否则，需要额外的逻辑
foreach($viewmodel as $item) {
$data[$item['place']['lat']=$item['lat'];
$data[$item['place']['long']=$item['lon'];
$data[$item['place']['schools'][]=$item['pname'];
}
如果使用php 5.3+，则可以使用lambda函数映射数组
$output = array();
$sort_schools = function($value, $key)
{
if ( ! is_array($output[$value['place'])
{
$output[$value['place'] = array();
}
if ( ! isset($output[$value['place']['lat'] && ! isset($output[$value['place']]['lon'])
{
$output[$value['place']]['lat'] = $value['lat'];
$output[$value['place']]['lon'] = $value['lon'];
}
$output[$value['place']]['schools'][] = $value['pname'];
};
array_map($sort_schools, $viewmodel);
或者，您可以在foreach循环或匿名函数中的lambda函数中使用类似的结构。
下面应该会产生所描述的预期结果
$arr = array(
array( 'place_id' => 1, 'place' => 'Berlin', 'lat' => 52.519, 'lon' => 13.406, 'id' => 1, 'pname' => 'Firstschool', 'typ' => 0, 's_id' => 32, 'fac' => 'history'),
array( 'place_id' => 1, 'place' => 'Berlin', 'lat' => 52.519, 'lon' => 13.406, 'id' => 1, 'pname' => 'Secondschool', 'typ' => 0, 's_id' => 32, 'fac' => 'history'),
array( 'place_id' => 1, 'place' => 'Berlin', 'lat' => 52.519, 'lon' => 13.406, 'id' => 1, 'pname' => 'Secondschool', 'typ' => 0, 's_id' => 32, 'fac' => 'history')
);
$result = array();
foreach($arr as $item) {
if (array_key_exists($item['place'], $result)) {
if (!in_array($item['pname'], $result[$item['place']]['schools'])) {
array_push($result[$item['place']]['schools'], $item['pname']);
}
} else {
$result[$item['place']]['lat'] = $item['lat'];
$result[$item['place']]['lon'] = $item['lon'];
$result[$item['place']]['schools'][] = $item['pname'];
}
}
print_r($result);
哪个应该输出
Array (
[Berlin] => Array
(
[lat] => 52.519
[lon] => 13.406
[schools] => Array
(
[0] => Firstschool
[1] => Secondschool
)
)
)