Php 试图创建数组但创建了一个对象

我目前正在尝试用php创建一个数组，该数组将采用当前格式： 日历[月份][行][列]，其中月份=1到6

这基本上就是我用来创建它的代码。还有很多，但我希望这足够了

function db_to_matrix($date_start, $date_end, $compliment, $workers, $worker_shifts, $shift_types)
{
    $calendar = array();
    //Sets the titles up for each months array
    $date = clone $date_start;
    while ($date <= $date_end)
    {
        $calendar[date_format($date,'n')] = array();
        $calendar[date_format($date,'n')][0][] = "db_table";
        $calendar[date_format($date,'n')][0][] = "id";
        $calendar[date_format($date,'n')][0][] = "";
        $calendar[date_format($date,'n')][1][] = "";
        $calendar[date_format($date,'n')][1][] = "";
        $calendar[date_format($date,'n')][1][] = "";
        $calendar[date_format($date,'n')][2][] = "compliment";
        $calendar[date_format($date,'n')][2][] = "compliment_day";
        $calendar[date_format($date,'n')][2][] = "Day Comp";
        $calendar[date_format($date,'n')][3][] = "compliment";
        $calendar[date_format($date,'n')][3][] = "compliment_day_actual";
        $calendar[date_format($date,'n')][3][] = "Act Day";
        $calendar[date_format($date,'n')][4][] = "compliment";
        $calendar[date_format($date,'n')][4][] = "compliment_night";
        $calendar[date_format($date,'n')][4][] = "Night Comp";
        $calendar[date_format($date,'n')][5][] = "compliment";
        $calendar[date_format($date,'n')][5][] = "compliment_night_actual";
        $calendar[date_format($date,'n')][5][] = "Act Night";
        date_modify($date,"+1 month");
    }
}

函数db\u to\u矩阵（$date\u start，$date\u end，$compaign，$workers，$worker\u shift，$shift\u types）
{
$calendar=array（）；
//设置每个月数组的标题
$date=克隆$date\u开始；
当（$date您可能正在重新使用变量时，请尝试使用其他变量名，或在while循环之前将$calendar重置为数组

$calendar=array（）；
为什么要创建索引为date的数组？例如，“date”可以等于1-6…也可以是3-6或6-12…我是PHP新手，所以可能完全偏离目标，但不应该将值传递给column元素？例如
$calendar[date\u format（$date，'n'）][0][**3**]=“db\u table”；
@Bendy:No.

[]添加了一个新元素。因此您将获得

[0][0]，[0][1]，[0][2]

@abracadver-谢谢，我会保持沉默；-）我会花钱，这就是本例的原因，因为我的PHP沙盒示例工作正常……这不起作用。我将上面的代码更改为显示。相同的输出。