我在谷歌上找到了一些解决方案,但它们没有满足我的要求。 我有一个数组,比较两个数组并存储在另一个数组中匹配和无与伦比的作为0
$current_week = self::CurrentWeekDateRange($s_date, $e_date);
它给我的结果是:
[0] => 2016-09-06
[1] => 2016-09-07
[2] => 2016-09-08
[3] => 2016-09-09
[4] => 2016-09-10
[5] => 2016-09-11
[6] => 2016-09-12
[7] => 2016-09-13
[8] => 2016-09-14
[9] => 2016-09-15
[10] => 2016-09-16
[11] => 2016-09-17
[12] => 2016-09-18
现在我的下一个数组是这样的:用户登录$返回一个阵列
[0] => Array
(
[date_log] => 2016-09-08
[total] => 15
)
[1] => Array
(
[date_log] => 2016-09-13
[total] => 30
)
[2] => Array
(
[date_log] => 2016-09-14
[total] => 400
)
不,我有3个用户表示它打印3次用户日志。 所以我想什么是我想要date_log符合我上面的第一阵列,
如果它匹配它会给它会得到存储在另一个数组,如果不匹配,那么它将存储0
我的问题我使用两个环路蚂蚁正在打印LOOP1 *循环2时代价值,但我想只有$ current_week时间值
我想是这样的:
$current_week = self::CurrentWeekDateRange($s_date, $e_date);
$i = 0;
foreach ($current_week as $day){
foreach ($returns as $return) {
if($day == $return['date_log']){
$array_total_hours[$i]['total'] = $return['total'];
$array_total_hours[$i]['date_log'] = $return['date_log'];
}
else {
$array_total_hours[$i]['date_log'] = $return['date_log'];
$array_total_hours[$i]['total'] = 0;
}
$i++;
}
}
print($array_total_hours);
我想我的结果是这样的:
[2016-09-06] => Array
(
[date_log] => 2016-09-06
[total] => 0
)
[2016-09-07] => Array
(
[date_log] => 2016-09-07
[total] => 30
)
[2016-09-08] => Array
(
[date_log] => 2016-09-08
[total] => 400
)
[2016-09-09] => Array
(
[date_log] => 2016-09-09
[total] => 0
)
.
.
.
.
.
.
.
[2016-09-18] => Array
(
[date_log] => 2016-09-18
[total] => 0
)
是的它工作完美; –
thx ...我更换了一些线条更清晰 - 我想你也这样做了。 – hummingBird