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我必须显示员工周六和周日的记录。我如何使用 php 来做到这一点。

while($row=mysql_fetch_row($res))
    { 
        list($name, $crd,$indate,$in,$out,$t) = $row;   
        $row[3] = date("g:i A", $in);
        $row[4] = date("g:i A" ,$out);
        $d3 = date("d-m-Y (D)", $in);
            $bg = '';
            if(date('D', $in) == 'Sun' || date('D', $in) == 'Sat')
                $bg = 'bgcolor=#EEEEEE';
                else 
                $bg = 'bgcolor= #ECE5B6';
                if ($ddate != $row[2] && $row[2] != null)
                {   
                    echo "<tr $bg><td colspan=4>Date: ".$d3."</td></tr>";
                    $ddate = $row[2];
                }
                else if($row[2]==null)
                {
                    $bg = 'bgcolor=#EEEEEE';
                    echo "<tr $bg><td colspan=4>Date: ".$from."</td></tr>";
                }

                if ($row[2] != null)
                {
                    echo "<tr>";
                    echo "<td align=left width='35%'>".trim($name)."</td>";
                    echo "<td align=center width='20%' >".$row[3]."</td>";
                    echo "<td align=center  width='23%' >".$row[4]."</td>";
                    echo "<td align=center width='16%'>".$t."</td>";
                    echo "</tr>";
                }

                else
                {
                    echo "<tr>";
                    echo "<td align=left width='35%'>".trim($name)."</td>";
                    echo "<td align=center width='20%' >Null</td>";
                    echo "<td align=center  width='23%' >Null</td>";
                    echo "<td align=center width='16%'>".$t."</td>";
                    echo "</tr>";
                }

            }

现在这个问题是星期天所有记录都显示没有日期。

mysql是

SELECT mstEmp.empname, mstEmp.cardno, InDate, 
dailyattendance.InTime,dailyattendance.OutTime, mstEmp.teamtype 
FROM mstEmp 
LEFT JOIN dailyattendance
 ON (dailyattendance.HolderName = mstEmp.empname )  
 AND (InDate BETWEEN '$from' AND '$to') 
 WHERE (mstEmp.teamtype = '$chk') 
order by InDate DESC

我有这个代码调用上面的代码

$sFrom = toYYYYMMDD($frm);
$sTo = toYYYYMMDD($t) . " 23:59:59";
while (strtotime($sFrom) <= strtotime($sTo)) 
{   
        setvalue($checkbox,$sFrom,$sFrom);
        $sFrom = date ("Y-m-d", strtotime("+1 day", strtotime($sFrom)));
 }
4

1 回答 1

0

您应该将 DATE 条件放在 WHERE 子句中

SELECT mstEmp.empname, mstEmp.cardno, InDate,  
dailyattendance.InTime,dailyattendance.OutTime, mstEmp.teamtype  
FROM mstEmp  
LEFT JOIN dailyattendance 
 ON (dailyattendance.HolderName = mstEmp.empname )   
 WHERE (mstEmp.teamtype = '$chk')  AND (InDate BETWEEN '$from' AND '$to')  
order by InDate DESC
于 2012-07-30T06:03:48.727 回答