我有一個名為 job counts 的數組，如下$job =>Array ( [0] => Array([count] => 3[yearmonth] => 2019-7) [1] => Array([count] => 3[yearmonth] => 2019-9) [2] => Array([count] => 5[yearmonth] => 2019-10)) )第二個日期數組為$date =>Array ([0] => Array([yearmonth] => 2019-6) [1] => Array([yearmonth] => 2019-7) [2] => Array([yearmonth] => 2019-8) [3] => Array([yearmonth] => 2019-9) [4] => Array([yearmonth] => 2019-10))作業數組不是連續數組。所以這里我想要的是，如果一個條目在 $date 中可用但在 $job 中不存在，則創建第三個數組，其中 count = 0 和yearmonth 值。像下面這樣的東西Array ([0] => Array([count] => 0[yearmonth] => 2019-6) [1] => Array([count] => 3[yearmonth] => 2019-7) [2] => Array([count] => 0[yearmonth] => 2019-8) [3] => Array([count] => 3[yearmonth] => 2019-9) [4] => Array([count] => 5[yearmonth] => 2019-10))
3 回答
楊魅力
TA貢獻1811條經驗 獲得超6個贊
$jobs要在迭代數組時有效地搜索數組中的日期$dates，請創建一個“查找”數組。
查找應該具有代表關系數據的鍵 ( yearmonth)。
使用數組中遇到的每個日期$dates，檢查該yearmonth值是否表示為查找中的鍵 - 如果是，則使用該count值，如果不是，則設置 0。
&變量之前的意思是“通過引用修改——這樣原始數組就會發生變化，而不需要聲明一個新的輸出數組。
??是“空合并運算符”，這允許在變量未聲明時使用后備值null。
代碼：（演示）
$jobs = [
['count' => 3, 'yearmonth' => '2019-7'],
['count' => 3, 'yearmonth' => '2019-9'],
['count' => 5, 'yearmonth' => '2019-10'],
];
$dates = [
['yearmonth' => '2019-6'],
['yearmonth' => '2019-7'],
['yearmonth' => '2019-8'],
['yearmonth' => '2019-9'],
['yearmonth' => '2019-10'],
];
$lookup = array_column($jobs, 'count', 'yearmonth');
foreach ($dates as &$date) {
$date['count'] = $lookup[$date['yearmonth']] ?? 0;
}
var_export($dates);
輸出：
array (
0 =>
array (
'yearmonth' => '2019-6',
'count' => 0,
),
1 =>
array (
'yearmonth' => '2019-7',
'count' => 3,
),
2 =>
array (
'yearmonth' => '2019-8',
'count' => 0,
),
3 =>
array (
'yearmonth' => '2019-9',
'count' => 3,
),
4 =>
array (
'yearmonth' => '2019-10',
'count' => 5,
),
)
或者，如果您要求聲明一個新的輸出數組，并且關聯子數組的順序必須與您的問題中的順序相同，那么這是我對相同技術的調整：
代碼：（演示）
$lookup = array_column($jobs, null, 'yearmonth');
$result = [];
foreach ($dates as $date) {
$result[] = $lookup[$date['yearmonth']] ?? ['count' => 0] + $date;
}
var_export($result);
查找數組現在包含完整的子數組數據。 array_column()用作null第二個參數來防止隔離單行，yearmonth用作第三個參數來分配第一級鍵。
foreach 循環不再“通過引用修改”?？蘸喜⑦\算符仍然用于避免調用isset()，并且非常適合簡潔地編寫后備數據。count我在包含元素的硬編碼數組和數組之間使用“聯合運算符” $date——避免調用array_merge()或手動編寫['yearmonth' => $date['yearmonth']. 這是一種合適的合并技術，因為鍵對于每個子數組來說都是唯一的且關聯的。
LEATH
TA貢獻1936條經驗 獲得超7個贊
永遠不要低估能夠在 php 數組中以非常靈活的方式/方式使用鍵的力量。
<?php
//Copied from mickmackusa
$jobs = [
['count' => 3, 'yearmonth' => '2019-7'],
['count' => 3, 'yearmonth' => '2019-9'],
['count' => 5, 'yearmonth' => '2019-10'],
];
$dates = [
['yearmonth' => '2019-6'],
['yearmonth' => '2019-7'],
['yearmonth' => '2019-8'],
['yearmonth' => '2019-9'],
['yearmonth' => '2019-10'],
];
//End copy
/*
Make the keys of $jobs be the same as values
of yearmonth: (This makes it easy to compare later on)
Array
(
[0] => Array
(
[count] => 3
[yearmonth] => 2019-7
)
[1] => Array
(
[count] => 3
[yearmonth] => 2019-9
)
[2] => Array
(
[count] => 5
[yearmonth] => 2019-10
)
[2019-7] => Array
(
[count] => 3
[yearmonth] => 2019-7
)
[2019-9] => Array
(
[count] => 3
[yearmonth] => 2019-9
)
[2019-10] => Array
(
[count] => 5
[yearmonth] => 2019-10
)
)
*/
foreach($jobs as $key=>$item) {
$jobs[$item['yearmonth']] = $item;
}
//Create a third array $third_arr based on your $jobs and $dates array
$third_arr = [];
foreach($dates as $item) {
$key_value = $item['yearmonth'];
if (isset($jobs[$key_value]['yearmonth'])) {
//Available in dates and present in $jobs
//Just copy values from the $jobs item which relates to this yearmonth
$third_arr[] = $jobs[$key_value];
}
else {
//Available in dates but not present in $job
$third_arr[] = ['count'=>0, 'yearmonth'=>$key_value];
}
}
第三個數組的輸出$third_arr：
Array
(
[0] => Array
(
[count] => 0
[yearmonth] => 2019-6
)
[1] => Array
(
[count] => 3
[yearmonth] => 2019-7
)
[2] => Array
(
[count] => 0
[yearmonth] => 2019-8
)
[3] => Array
(
[count] => 3
[yearmonth] => 2019-9
)
[4] => Array
(
[count] => 5
[yearmonth] => 2019-10
)
)
上面代碼的更壓縮版本如下所示：
foreach($jobs as $key=>$item) {
$jobs[$item['yearmonth']] = $item;
}
$third_arr = [];
foreach($dates as $item) {
$kvalue = $item['yearmonth'];
isset($jobs[$kvalue]['yearmonth']) ?
$third_arr[] = $jobs[$kvalue] : $third_arr[] = ['count'=>0, 'yearmonth'=>$kvalue];
}
Qyouu
TA貢獻1786條經驗 獲得超11個贊
因此，如果您添加count => 0到$date數組和索引，yearmonth您可以索引$job并將yearmonth其合并到$date：
$result = array_merge(array_column(array_map(function($v) { $v['count'] = 0; return $v; },
$date), null, 'yearmonth'),
array_column($job, null, 'yearmonth'));
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