Question

我知道在更改数组中的键之前已经问过这个问题，但是如果密钥中的日期与日期范围相匹配，我想要的是那个密钥应该被替换。

E-克我有一个数组，其中包含从第1天到第31天的整个月的按键

[Project 456] => Array
    (
        [2014-12-01] => 5
        [2014-12-02] => 00
        [2014-12-03] => 00
        [2014-12-04] => 00
        [2014-12-05] => 00
        [2014-12-06] => 00
        [2014-12-07] => 00
        [2014-12-08] => 00
        [2014-12-09] => 00
        [2014-12-10] => 8
        [2014-12-11] => 00
        [2014-12-12] => 00
        [2014-12-13] => 00
        [2014-12-14] => 00
        [2014-12-15] => 00
        [2014-12-16] => 00
        [2014-12-17] => 00
        [2014-12-18] => 00
        [2014-12-19] => 00
        [2014-12-20] => 00
        [2014-12-21] => 00
        [2014-12-22] => 00
        [2014-12-23] => 00
        [2014-12-24] => 00
        [2014-12-25] => 00
        [2014-12-26] => 00
        [2014-12-27] => 00
        [2014-12-28] => 00
        [2014-12-29] => 00
        [2014-12-30] => 00
        [2014-12-31] => 00
        [Total] => 13
    )

我希望日期与其他数组的日期范围相匹配

Array
(
    [0] => stdClass Object
        (
            [AvailableLeaveTypes] => Sick
            [TotalAllocatedLeaves] => 6
            [TotalLeavesTaken] => 3
            [Balance] => 3
            [LeaveFrom] => 2014-12-05
            [LeaveTo] => 2014-12-08
        )
)

然后具有匹配日期的第一个数组的键应替换为LeaveType，即Sick

我必须遵循哪种方法？

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更新

First Array实际上是MultiDimentional Array。

这是完整的数组

我想如果我可以使用IF语句，如果Top Arrays Keys不是Day和Date那么可以匹配内部日期。所以比较两个数组的可能结果就是。

但我不想打扰它所处的顺序。只想改变键。

这是另一个完整数组，但其中只有一个拥有日期范围的数据。

Array
(
    [0] => stdClass Object
        (
            [AvailableLeaveTypes] => Sick
            [TotalAllocatedLeaves] => 6
            [TotalLeavesTaken] => 3
            [Balance] => 3
            [LeaveFrom] => 2014-12-05
            [LeaveTo] => 2014-12-08
        )

    [1] => stdClass Object
        (
            [AvailableLeaveTypes] => On Leave
            [TotalAllocatedLeaves] => 0
            [TotalLeavesTaken] => 0
            [Balance] => 0
            [LeaveFrom] => 
            [LeaveTo] => 
        )

    [2] => stdClass Object
        (
            [AvailableLeaveTypes] => Earned Leave
            [TotalAllocatedLeaves] => 0
            [TotalLeavesTaken] => 0
            [Balance] => 0
            [LeaveFrom] => 
            [LeaveTo] => 
        )

    [3] => stdClass Object
        (
            [AvailableLeaveTypes] => Casual
            [TotalAllocatedLeaves] => 0
            [TotalLeavesTaken] => 0
            [Balance] => 0
            [LeaveFrom] => 
            [LeaveTo] => 
        )

    [4] => stdClass Object
        (
            [AvailableLeaveTypes] => Annual
            [TotalAllocatedLeaves] => 0
            [TotalLeavesTaken] => 0
            [Balance] => 0
            [LeaveFrom] => 
            [LeaveTo] => 
        )

    [5] => stdClass Object
        (
            [AvailableLeaveTypes] => Unpaid
            [TotalAllocatedLeaves] => 0
            [TotalLeavesTaken] => 0
            [Balance] => 0
            [LeaveFrom] => 
            [LeaveTo] => 
        )

)

因此，如果LeaveFrom和LeaveTo为空，则不会为其更新第一个数组。只适用于那些有来自/到日期的人。

Answer 1

你可以使用一个简单的foreach循环。首先，使用离开的空闲对象清理假数组，因为您不需要处理它们：

$leaves_array = array_filter($leaves_array, function($val){
    return (!empty($val->LeaveFrom) && !empty($val->LeaveTo));
});

然后在循环上，父循环是叶子，内循环是项目。

尝试将它们转换为DateTime对象，以便您可以轻松地比较日期。然后在处理过程中，如果该键位于日期范围内，则将其放入临时持有者中，然后取消设置。取消设置后，再按下另一组新密钥，然后按键：

foreach($leaves_array as $leave) {
    $range = array();
    foreach ($project_array['Project 456'] as $date => $value) {
        try {
            $new_date = new DateTime($date);
            $from = new DateTime($leave->LeaveFrom);
            $to = new DateTime($leave->LeaveTo);
            if($new_date >= $from && $new_date <= $to) {
                $range[] = $date . '-Sick';
                unset($project_array['Project 456'][$date]);
            }
        } catch(Exception $e) {

        }

    }
    // changes
    foreach($range as $range_date) {
        $project_array['Project 456'][$range_date] = '00';
    }
    // resort
    ksort($project_array['Project 456']);
}

Sample Output

但我的建议是，不是改变这些键，为什么不在这些日期内创建一个数组，然后只是在其中推送相关的细节，比如留下日期和东西。

如果Key中的Date与其他数组的日期范围匹配，则更改PHP数组中的键名称

1 个答案: