我正在运行一个cron作业,将任务日期更改为当前一周。目前,我正在更改日期,采取日期和移动它在一个星期。
当前主计长代码:
function weekly_tasks()
{
$this->load->model('tasksmodel');
$data['weekly_tasks'] = $this->tasksmodel->get_weekly_tasks();
var_dump($data);
foreach ($data['weekly_tasks'] as $row)
{
// change parent task date to next week
$date = $row->due_date;
$newdate = strtotime ( '+1 week' , strtotime ( $date ) ) ;
$newdate = date ( 'Y-m-j' , $newdate );
$row->due_date = $newdate;
// set up task update var
$task_update = $row;
$task_id = $row->task_id;
//create this week's task
$task_create = $row;
$this->tasksmodel->create_weekly_tasks($task_create);
var_dump($task_create);
// update the task to next week
$this->tasksmodel->update_weekly_tasks($task_update, $task_id);
var_dump($task_update);
}
}上述代码的日期为2015/10/14周,并将该日期更改为同一工作日,即2015/10/21周。
现在,我需要在过去的任何日期,并将其更改为同一天的当前一周,在原来的日期。因此,2015/09/09应改为2015/10/21这周的同一天。
我是否仍然通过获取旧日期和今天之间的间隔间隔来使用间隔,还是有更好的方法来做到这一点?
发布于 2015-10-19 06:36:03
我建议不要使用字符串,而应该使用DateTime类:
$date = new Date($row->due_date);
$newdate = new Date(); // this is obviously in the "current" week.
// get the week day of the "old" and "new" date:
$oldWD = $date->format("N");
$newWD = $newdate->format("N");
// set current weeks weekday as diff of weekday:
$diff = $oldWD - $newWD;
if ( $diff > 0 )
$newdate->add( new DateInterval("P" . $diff . "D") );
else if ( $diff < 0 ) // $diff == 0 means no changes
$newdate->sub( new DateInterval("P" . abs($diff) . "D") );
// adjust times to the time of the old date
$newdate->setTime(date->format("H"), date->format("i"), date->format("s"));(现在不要对$diff进行正确的计算,而不是测试,可能是相反的)
https://stackoverflow.com/questions/33207399
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