Question

我有一段时间（旅行），例如2015-04-01 - 2015-04-23。

现在，还有其他几个时期（条件），例如： 2015-03-27 - 2015-04-02
2015-04-04 - 2015-04-18
2015-04-02 - 2015-04-06
2015-04-14 - 2015-04-16
2015-04-23 - 2015-04-30

这里是草图（黑色是旅行，红色是条件期间） enter image description here

目标是在没有间隙（如果可能），重叠和重复的情况下覆盖整个旅程。

所以我的最终条件是这样的：
2015-04-01 - 2015-04-02
2015-04-02 - 2015-04-03
2015-04-03 - 2015-04-18
2015-04-23 - 2015-04-23

新草图： enter image description here

当我得到条件时，条件没有特定的顺序。

我目前的实施非常庞大且有缺陷。有没有一种行之有效的方法呢？

请注意，期间日期是示例，草图并不准确。

Answer 1

您好，这是＆＃34; quck ＆＃34;演示...... 我希望能够清楚地理解。

您的约会对象：

$travel = array('from'=>'20150401', 'to'=>'20150423');
$conditions = array(
    'condition_1' => array('from'=>'20150327', 'to'=>'20150402'),
    'condition_2' => array('from'=>'20150404', 'to'=>'20150418'),
    'condition_3' => array('from'=>'20150402', 'to'=>'20150406'),
    'condition_4' => array('from'=>'20150414', 'to'=>'20150416'),
    'condition_5' => array('from'=>'20150423', 'to'=>'20150430')
);

准备日期的功能：

// CONVERTS DATES INTO TIME
function get_times($from, $to){
    preg_match('/([0-9]{4})([0-9]{2})([0-9]{2})/', preg_replace("/\D/m", '', $from), $d); $from = @mktime(0, 0, 0, $d[2], $d[3], $d[1]);
    preg_match('/([0-9]{4})([0-9]{2})([0-9]{2})/', preg_replace("/\D/m", '', $to), $d); $to = @mktime(0, 0, 0, $d[2], $d[3], $d[1]);
    $delta = $to-$from;
    return array('from'=>$from, 'to'=>$to, 'delta'=>$delta, 'days'=>$delta/86400);
}

// PARSE MORE DATA INTO $travel & $conditions
function parse_dates($travel, $conditions){
    $travel['times'] = get_times($travel['from'], $travel['to']);
    $applicable_conditions = array();
    foreach($conditions as $i=>$cond){
        if($cond['from']<=$travel['to'] && $cond['to']>=$travel['from']){
            $cond['times'] = get_times($cond['from']<$travel['from']?$travel['from']:$cond['from'], $cond['to']>$travel['to']?$travel['to']:$cond['to']);
            $applicable_conditions[$i] = $cond;
        }
    }
    return array('travel'=>$travel, 'conditions'=>$applicable_conditions);
}

举个例子，你可能想把它变成更适合你的东西：

// (mockup) DISPLAY GRAPHICS
function show_graph($dates){
    $w = 600; $w2 = 100; $h = 30; $bg = '#4682B4'; $col = '#FFFFFF';
    $day = '<div style="display:inline-block; width:'.(round($w/$dates['travel']['times']['days'])-2).'px; border:1px solid '.$col.'; height:'.($h-2).'px;';
    $day_blue = $day.' background-color:'.$bg.';">';
    $day_red = $day.' background-color:red;">';
    $day_empty = $day.' background-color:white;">';
    $day_black = $day.' background-color:black;">';
    $cd = '</div>';
    $days = ''; $cond_days = $day_nocond = array();
    for($i=1; $i<=$dates['travel']['times']['days']; $i++){
        $daytime = $dates['travel']['times']['from']+(86400*$i);
        $days .= $day_blue.$i.$cd;
        $day_havecond[$i] = false;
        foreach($dates['conditions'] as $name=>$cond){
            $is_cond = $daytime>=$cond['times']['from'] && $daytime<=$cond['times']['to'];
            $day_box = $is_cond ? ( $day_havecond[$i] ? $day_black.$i.$cd : $day_red.$i.$cd ) : $day_empty.'&nbsp;'.$cd;
            if($is_cond && !$day_havecond[$i]){ $day_havecond[$i] = true;}
            @$cond_days[$name] .= $day_box;
        }
    }
    $o = '
        <div style="display:inline-block; width:'.$w2.'px;">Travel</div>
        <div style="display:inline-block; color:'.$col.';">'.$days.'</div><br/>';
    foreach($cond_days as $name=>$days){
        $o .= '
        <div style="display:inline-block; width:'.$w2.'px;">'.$name.'</div>
        <div style="display:inline-block; color:'.$col.';">'.$days.'</div><br/>';
    }
    return '<div style="width:'.($w+$w2).'px;">'.$o.'</div>' ;
}

这就是你制作图形的方法：

# USAGE
$dates = parse_dates($travel, $conditions);
echo show_graph($dates);

# DEBUG
echo '<hr>'; print_r($dates);

涵盖多个（较小）期间的期间

1 个答案: