我不确定我做错了什么，但基本上我试图通过数组（done）列出一个月中的日期，但trips表中预订日期范围内的日期对它们来说有不同的外观，这就是目前的样子
http://oi60.tinypic.com/2zzrc03.jpg
我正试图这样输出。。
March 1[2 3 4 5]6[7 8 9 10]
代码在很大程度上是有效的，但对于有多个事件的月份，它会重复该月份，并在自己的月份行中显示日期
这是我的代码，是的，我知道它是mysql，但目前这是一个本地项目，所以我正在努力让它在进入这一步之前发挥作用。
<table width="100%" cellspacing="0">
<?php
$cmonth = date('F');
$cyear = date('Y');
$sql = "SELECT * FROM calendar WHERE year = '$cyear' ORDER BY m_order ASC";
$res = mysql_query($sql);
while ($rows = mysql_fetch_array($res)) {
$month_end = $rows['days_in_month'];
$month_name = $rows['month_name'];
$m_order = $rows['m_order'];
$sql2 = "SELECT * FROM trips WHERE start_date LIKE '____-0$m_order-__' ORDER BY start_date ASC";
$res2 = mysql_query($sql2);
while ($row = mysql_fetch_array($res2)) {
$stdate = $row['start_date'];
$s = date_parse_from_format("Y-m-d", $stdate);
$endate = $row['end_date'];
$e = date_parse_from_format("Y-m-d", $endate);
$start = $s['day'];
$end = $e['day'];
?>
<tr>
<td width="80px"><?php echo $month_name; ?></td>
<?php
foreach(range(1, $month_end) as $days)
{
if(in_array($days, range($start, $end)))
{
echo "<td style='"background-color: #ccc;'" align='"center'">". $days ." </td>";`
}
else
echo "<td align='"center'">". $days ."</td>";
}
?>
</tr>
<?php }
} ?>
</table>
我只是想知道我的编码是否错误，或者我是否应该尝试用另一种方法来实现我的目标？欢迎提出任何建议。
通过四处玩耍和与人交谈，我使它发挥了作用，消除了对日历表的需求，并更改了数组。
<?php
function date_range($first, $last, $step = '+1 day', $output_format = 'Y-m-d' ) {
$dates = array();
$current = strtotime($first);
$last = strtotime($last);
while( $current <= $last ) {
$dates[] = date($output_format, $current);
$current = strtotime($step, $current);
}
return $dates;
}
?>
<table width="100%" cellspacing="0" cellpadding="0">
<?php
$cmonth = date('F');
$cyear = date('Y');
$sql = "SELECT * FROM trips WHERE year(start_date) = '$cyear' ORDER BY start_date ASC";
$res = mysql_query($sql);
$array_days = array();
while ($rows = mysql_fetch_array($res)) {
$all_dates[] = array('start'=>strtotime($rows['start_date']),'end'=>strtotime($rows['end_date']));
$selected_dates = date_range(date("Y-m-d",strtotime($rows['start_date'])), date("Y-m-d",strtotime($rows['end_date'])));
foreach($selected_dates as $selected_date){
$array_days[$selected_date] = date("d",strtotime($selected_date));
}
}
$current_month = date("m");
$next_6_month = date("m", strtotime("+5 month", strtotime(date("F") . "1")));
for($i=$current_month;$i<=$next_6_month;$i++){ // 12 months in year
?>
<tr>
<td width="40px"><?php echo date('M', mktime(0, 0, 0, $i,10, $cyear)); ?></td>
<?php
$days_in_month = cal_days_in_month(CAL_GREGORIAN,$i,$cyear);
foreach(range(1, $days_in_month) as $days)
{
if(array_key_exists(date('Y-m-d', mktime(0, 0, 0, $i,$days, $cyear)),$array_days)){
echo "<td style='text-align:center;background-color:ccc' width='12px'>$days</td>";
}else{
echo "<td style='text-align:center;' width='12px'>$days</td>";
}
}
?>
</tr>
<?php } ?>
</table>
首先使用prepared语句和mysqli，因为使用当前方法可能会注入sql！
尝试使用SELECT DISTINCT
http://www.w3schools.com/sql/sql_distinct.asp