我编写了以下代码,在PHP中获取数组,并像在MySQL中使用LEFT join一样将它们"连接"在一起。为了我的缘故,我使用foreach编写了函数,并将数组传入并返回了一个新数组。。。我觉得它很可读,但我也知道它效率低下。
理想情况下,我希望这个函数能处理每个数组10000行以上的行,我怀疑这意味着:1)通过引用传递$original(以节省内存),2)使用一个PHP迭代数组函数而不是foreach(以节省处理时间)。
通常我不会发布这些类似"测验"的问题,但我觉得这个问题的答案会让社区受益。(就像这个家伙:加入2个多维数组)
愿最快、消耗内存最少的答案获胜!:P
<?php
// Join Arrays on Keys (**updated with knittl's suggestion**)
function array_join($original, $merge, $on) {
if (!is_array($on)) $on = array($on);
foreach ($merge as $remove => $right) {
foreach ($original as $index => $left) {
foreach ($on as $from_key => $to_key) {
if (!isset($original[$index][$from_key])
|| !isset($right[$to_key])
|| $original[$index][$from_key] != $right[$to_key])
continue 2;
}
$original[$index] = array_merge($left, $right);
unset($merge[$remove]);
}
}
return array_merge($original, $merge);
}
// Test Arrays
$data1 = array(
array(
'productId' => '822335',
'dateHour' => '2011-11-17 06:00:00',
'qtySold' => '200',
'qtyCanceled' => '10',
),
array(
'productId' => '822335',
'dateHour' => '2011-11-17 07:00:00',
'qtySold' => '100',
'qtyCanceled' => '20',
),
array(
'productId' => '822336',
'dateHour' => '2011-11-17 06:00:00',
'qtySold' => '0',
'qtyCanceled' => '30',
),
array(
'productId' => '822336',
'dateHour' => '2011-11-17 07:00:00',
'qtySold' => '50',
'qtyCanceled' => '40',
),
);
$data2 = array(
array(
'entity_id' => '822335',
'dateHour' => '2011-11-17 06:00:00',
'productInventory' => '300',
),
array(
'entity_id' => '822335',
'dateHour' => '2011-11-17 07:00:00',
'productInventory' => '200',
),
array(
'entity_id' => '822336',
'dateHour' => '2011-11-17 06:00:00',
'productInventory' => '100',
),
array(
'entity_id' => '822336',
'dateHour' => '2011-11-17 07:00:00',
'productInventory' => '50',
),
);
// Usage
$result = array_join($data1, $data2, array(
'productId' => 'entity_id',
'dateHour' => 'dateHour'
));
print_r($result);
更好?
function array_join($original, $merge, $on) {
if (!is_array($on)) $on = array($on);
foreach ($merge as $remove => $right) {
foreach ($original as $index => $left) {
foreach ($on as $from_key => $to_key) {
if (!isset($original[$index][$from_key])
|| !isset($right[$to_key])
|| $original[$index][$from_key] != $right[$to_key])
continue 2;
}
$original[$index] = array_merge($left, $right);
unset($merge[$remove]);
}
}
return array_merge($original, $merge);
}