php - 加速 PHP 脚本
问题描述
我正在为我的办公室制定日历。每个人都有自己的专栏,每天都有一条线。
有一些定期日期,例如,考虑到人们必须在周末工作。但大多数日期都来自 MySQL 数据库。
所以它做了一个双重循环(人与日期),其中每个日期都必须检查这个人,他在这一天有什么样的职业。
有没有办法优化这个脚本,因为在线上至少需要2-3秒(根据PHP只有0.03秒,但我觉得它不正确)和超过8秒(再次根据PHP)在我们的网络!这只是 5 个月,我们想拥有它一整年。
您可以在此处找到测试版本(仅查看 HTML 和 CSS): http: //mybees.ch/for/tableau.php
这是其中的PHP:
//Creating the line for the collaborators
$ids;
$ma_count=0;
$ma_name_query = mysqli_query($bdi,"SELECT DISTINCT m.id, m.name, m.vorname, m.grup FROM ma m, ma_pos mp WHERE m.id = mp.maid AND m.grup != 14 ORDER BY m.grup, mp.posid, m.name, m.vorname");
while($ma_name = mysqli_fetch_assoc($ma_name_query)) {
echo '<div class="cell name">'.$ma_name['name'].' '.$ma_name['vorname'].'</div>';
$ids[$ma_count] = $ma_name['id'];
$grup[$ma_count] = $ma_name['grup'];
$ma_count++;
}
// Check if special group day
$firstGroup = 1;
$firstDate = strtotime("$year-01-01 00:00 GMT");
$picket = array();
if (date("N", $firstDate) == 2)// Tuesday
$firstDate -= 24 * 3600;
elseif (date("N", $firstDate) == 3)// Wednesday
$firstDate -= 2 * 24 * 3600;
elseif (date("N", $firstDate) == 5)// Friday
$firstDate -= 24 * 3600;
elseif (date("N", $firstDate) == 6)// Saturday
$firstDate -= 2 * 24 * 3600;
elseif (date("N", $firstDate) == 7)// Sunday
$firstDate -= 3 * 24 * 3600;
for ($date = $firstDate; $date <= strtotime("$year-12-31 00:00 GMT"); $date += 24 * 3600) {
$weekNb = date("W",$date);
$weekDay = date("N",$date); // Monday = 1, Sunday = 7
if ($weekDay < 4)
$group = $weekNb % 4 - 1 + $firstGroup;
else
$group = $weekNb % 4 - 2 + $firstGroup;
if ($group == 0)
$group = 4;
if ($group == -1)
$group = 3;
$picket[$date] = $group;
}
$groupColor = ["yellow", "blue", "red", "green"];
function isPicket($date, $grup) {
global $picket, $groupColor;
//global $picket, $groupColor;
if ($grup < 5) {
if ($picket[$date] == $grup)
return " style='background-color: ".$groupColor[$picket[$date]-1]."'";
}
}
$today_stp = time() - time() % (24 * 3600) - 11 * 24 * 3600;
$frei_query_text = "SELECT id_ma, date1, date2, free.id as type, moment, remark, location FROM frei, free WHERE free.id = frei.type AND date1 BETWEEN CAST('$date1' AS DATE) AND CAST('$date2' AS DATE)";
$frei_query = mysqli_query($bdi, $frei_query_text);
$free = array();
while($frei = mysqli_fetch_assoc($frei_query)) {
$free[] = $frei;
}
// Filling the lines
for ($date = strtotime($date1); $date <= strtotime($date2); $date += 24 * 3600) {
$today = ($date == $today_stp) ? ' id="today"':'';
echo "<div class='clear'$today>";
$class = (date('N', $date) < 6) ? 'week' : 'weekend';
echo "<div class='date $class line'>".date("D, d.m.", $date)."</div>";
for ($i = 0; $i < $ma_count; $i++) {
echo "<div class='cell line $class'".isPicket($date,$grup[$i]).">
<div class='small-cell".isColor($ids[$i], $date, 0)."' id='divUp-".$ids[$i]."-$date'> </div>
<div class='small-cell".isColor($ids[$i], $date, 1)."' id='divDown-".$ids[$i]."-$date'> </div>
</div>";
}
echo '</div>';
}
function isColor($id_ma, $date, $moment) {
global $free;
for ($i = 0; $i < count($free); $i++) {
if ($id_ma == $free[$i]['id_ma']) {
if ($date >= strtotime($free[$i]['date1']) && $date <= strtotime($free[$i]['date2'])) {
if ($free[$i]['moment'] == $moment || $free[$i]['moment'] == 2) {
$type = $free[$i]['type'];
$style = "";
if ($type > 1 && $type < 5)
$style = " urlaub";
if ($type > 4 && $type < 8)
$style = " frei";
if ($type > 7 && $type < 11)
$style = " ferien";
if (($type > 22 && $type < 34) || ($type > 37 && $type < 48))
$style = " kurse";
if ($type > 10 && $type < 17)
$style = " krank";
return " $style' title='".$free[$i]['remark'];
}
}
}
}
}
非常感谢您的帮助
解决方案
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