Question

嗨，对于stackoverflow中的所有人，我想请求你们的帮助。我一直坚持我正在做的事情，我想计算我的查询中存在的天数，但我不知道如何。有人可以帮我吗？非常感谢一种形式的帮助。

以下是我的评论代码：

//Used for getting the values               
$mfrom = $this->input->get('mfroma');  //From: Month
$yfrom = $this->input->get('yfroma');  //From: Year
$yto = $this->input->get('ytoa');      //To : Year
$n_of_months = $this->input->get('n_of_monthsa'); // Month or year Range

$cmonth = $mfrom;            //From: Month
$cyear = $yfrom;            //From: Year
$clastyear = $yto;          //To: Year

$sSQLSelect = "";
$sSQLSelect1 = "";

for($i=0;$i<=$n_of_months;$i++)
{
    $curdate = $cmonth." ".$cyear;

    $sSQLSelect = $sSQLSelect."SUM(IF(DATE_FORMAT(date_d, '%c %Y') = '$curdate' ,
                                    IF(DP.hrsdiff <= 4, 0.5, 1), 0)) AS 'month$cmonth$cyear', ";
    $sSQLSelect1 = $sSQLSelect1."month$cmonth$cyear + ";

    if($cmonth+1 > 12) {
        $cyear++;
        $cmonth = 1;
    } else
        $cmonth++;
}

date_default_timezone_set('Asia/Manila');
$current_date = date('Y-m',time());

$d = array();         
$i = 0;

$sSQL = "SELECT TS.id AS TSid, employeeid, `date` AS date_d, TS.TI1 AS TI1, TS.TO1 AS TO1,
                    CONCAT(E.lastName,', ',E.firstName,' ',E.middleInitial,'.') AS 'fullname', 
                    (TIME_TO_SEC(CAST(TS.`TI1` AS time))/3600) AS 'timeinint',                  
                    (TIME_TO_SEC(CAST(TS.`TO1` AS time))/3600) AS 'timeoutint',
                    E.startDate, C.companyName, B.branchName,
                    R.mon, R.tue, R.wed, R.thu, R.fri, R.sat, R.sun
            FROM timeschedule AS TS
            LEFT JOIN  employee AS E ON TS.employeeid = E.id
            LEFT JOIN  company AS C ON E.companyid = C.id
            LEFT JOIN  branch AS B ON E.branchid = B.id
            LEFT JOIN restday AS R ON R.id = E.restDayid
            WHERE TS.`show` = '1'  $namecondition $companynamecondition $branchnamecondition 
            GROUP BY employeeid
            ORDER BY fullname ASC";

$result = $this->db->query($sSQL);
$data = array();

foreach($result->result() as $row) 
{   
    $rowemployeedetails = $row;
    $fullname = $row->fullname;
    $startdate = $row->startDate;


    $st_month = (int)(date("m",strtotime($startdate)));
    $st_day = (int)(date("d",strtotime($startdate)));
    $st_year = (int)(date("Y",strtotime($startdate)));
    $data['data'][$i][] = array($row->fullname);
    $data['data'][$i][] = array($row->companyName);
    $data['data'][$i][] = array($row->branchName);


    // #$cmonth = start_month
    // #$cyear = start_year
    // #number months in between
    $cmonth = $mfrom;
    $cyear = $yfrom;

    for($j=0; $j<= $n_of_months ;$j++){
        if((date('Y-m', strtotime($startdate)) <= date('Y-m', strtotime($cyear.'-'.$cmonth)))&&($current_date >= date('Y-m', strtotime($cyear.'-'.$cmonth))))
        {   
            $d[$j]  = ((date('Y-m', strtotime($cyear.'-'.$cmonth)) ==  date('Y-m',time()))?date('d', time()):cal_days_in_month(CAL_GREGORIAN, $cmonth, $cyear));
        }
        else
        {
            $d[$j]=0;
        }

        if($cmonth === 12){
            $cyear++;
            $cmonth = 1;
        }
        else{
            $cmonth++;
        }
    }

    // #Result: d[] = number of days in month
    // #$cyear = start year filter      
    // #$clastyear = last year filter
    // #$cmonth = start_month
    $cyear = $yfrom;
    $clastyear = $yto;


}
$result->free_result();
return json_encode($data);
}

Answer 1

如果您有两个日期并想知道从一个日期到另一个日期的天数，一个简单的解决方案就是使用DateTime::diff，如下所示：

<?php
$from = new DateTime('2014-03-28');
$to = new DateTime('2014-03-30');
$interval = $from->diff($to);
echo $interval->format('%R%a'); // +2

想用php找到现在的天数？

1 个答案: