在PHP / MySQL中使用Week（Date）时，如何将星期的第一天设置为星期一？

Question

我正在使用以下代码构建Google Chart，以便在一年内对应于Week Numbers的MySQL表中提取所有条目。 目前周数从周日开始，我想改变这一点，以便他们在星期一开始，但我不知道如何做到这一点。

$i=1;
    while($i<=53)
       {
        $week_start = new DateTime();
        $week_start->setISODate(2013,$i+1);
        $date_display = $week_start->format('j M Y');

        $sessions = $wpdb->get_var($wpdb->prepare("SELECT Sum(Due) from patient_sessions WHERE Type='Session' AND Week(Date)='$i'"));

        $temp = array();

        $temp[] = array('v' => (string) $date_display); 
        $temp[] = array('v' => (string) $sessions); 
        $rows[] = array('c' => $temp);
        $i++;
    }

修订后的代码

$sessions = $wpdb->get_results($wpdb->prepare("SELECT Due,Date from patient_sessions WHERE Type='Session'"));
$work_times = $wpdb->get_results($wpdb->prepare("SELECT Amount,Date from work_times"));
$expenses = $wpdb->get_results($wpdb->prepare("SELECT Amount,Date from expenses WHERE Client='Psychotherapy'"));

$i=1;
while($i<=53) {
    $week_start = new DateTime();
    $week_start->setISODate(2013,$i+1);
    $date_display = $week_start->format('j M Y');

    $session_total = 0;
    $work_time_total = 0;
    $expense_total = 0;


      foreach ($sessions as $session) {
    if (date("W", strtotime($session->Date)) == $i) {
    $session_total = ($session_total+$session->Due);
    }
    }

         foreach ($work_times as $work_time) {
    if (date("W", strtotime($work_time->Date)) == $i) {
    $work_time_total = ($work_time_total+$work_time->Amount);
    }
    }

          foreach ($expenses as $expense) {
    if (date("W", strtotime($expense->Date)) == $i) {
    $expense_total = ($expense_total+$expense->Amount);
    }
          }

    $balance = ($session_total + $work_time_total - $expense_total);

    $temp = array();

    $temp[] = array('v' => (string) $date_display); 
    $temp[] = array('v' => (string) $balance); 
    $rows[] = array('c' => $temp);
    $i++;
}

Answer 1

WEEK(date[mode]); 

date = a date value.    
mode = An integer indicating the starting of the week. 

默认的arugment是0，这是星期日，将其设置为1将是星期一，2星期二，依此类推。

week(date,1);

Answer 2

要使它在您的代码中工作，您应该能够将查询更改为：

"SELECT Sum(Due) FROM patient_sessions WHERE Type='Session' AND Week(Date,1)=$i"

但是，我想我可以更进一步，使用GROUP BY进行一次SQL调用，然后迭代结果;

"SELECT Week(Date,1), SUM(DUE) FROM patient_sessions WHERE Type='Session' GROUP BY Week(Date,1)"

或者，您可能需要使用WEEK（日期，2），具体取决于您希望如何处理跨越年份的周数。