最好是分两个字段来处理,starttime,endtime
如果非要这样,也可以用:
select * from table1 where cDate(left(rs("MM-DD"),5))>cDate("9.25") and cDate(right(rs("MM-DD"),5))<cDate("11.30")
如果非要这样,也可以用:
select * from table1 where cDate(left(rs("MM-DD"),5))>cDate("9.25") and cDate(right(rs("MM-DD"),5))<cDate("11.30")
数组形式是$a[$i]['start_dt']['end_dt'];
static $p = 0;
$retval = false;
if($p == 0) {
$p = $ar['end_dt'];
return $retval;
}
if($p < $ar['start_dt'])
$retval = array('start_dt' => $p+0.01, 'end_dt' => $ar['start_dt']-0.01 );
$p = $ar['end_dt'];
return $retval;
}foreach($a as $v) {
if($r = foo($v))
$ar[] = $r;
}
print_r($ar);计算是按十进制数计算的,因为你给的是十进制数(9.25)。
若是60进制的时间请自行转换
function m_d($v,$d) {
$d = array(31,31,28,31,30,31,30,31,31,30,31,30,31);
$t = split("\.",$v);
if($t[1] == $d[$t[0]] && $d>0) {
$t[0]++;
$t[1] = 1;
}else if($t[1] == 1 && $d<0) {
$t[0]--;
$t[1] = $d[$t[0]];
if($t[0] == 0) $t[0] = 12;
}else
$t[1] += $d;
return sprintf("%02d.%02d",$t[0],$t[1]);
}把原
$retval = array('start_dt' => $p+0.01, 'end_dt' => $ar['start_dt']-0.01 );
改做
$retval = array('start_dt' => m_d($p,1), 'end_dt' => m_d($ar['start_dt'],-1) );由于数据中没有给出年信息,所以未做闰年判断