Php 我想在日历中添加日期名称

我想用$day_num加上$day_of_week，但我得到了一些奇怪的组合（1周二2 Tuf 3拖船4 Tuh 5 Tui 6 Tuj 7 Tuk 8 Tul 9 Tum 10 Tun 11 Tuo 12 Tup）

日历上不再显示1 Tue 2 Tuf 3拖船等。我把它拿出来了

我希望日历看起来像这样（1周一2周二3周三4周四5周五6周六7周日）

这是导致问题的代码部分

//count up the days, untill we've done all of them in the month
 while ( $day_num <= $days_in_month )
 { 
 echo "<td> $day_num $day_of_week </td>"; 
 $day_num++; 
 $day_count++;
$day_of_week++;
 //Make sure we start a new row every week
 if ($day_count > 7)
 {
 echo "</tr><tr>";
 $day_count = 1;
 }
 }  

//计算天数，直到我们在一个月内完成所有工作
而（$day_num 7）
{
回声“；
$day_count=1；
}
}  

这是完整的代码，没有上面的修改

<?php 
 //This gets today's date 
 $date =time () ; 
 //This puts the day, month, and year in seperate variables 
 $day = date('d', $date) ; 
 $month = date('m', $date) ; 
 $year = date('Y', $date) ;
 //Here we generate the first day of the month 
 $first_day = mktime(0,0,0,$month, 1, $year) ; 
 //This gets us the month name 
 $title = date('F', $first_day) ;
 //Here we find out what day of the week the first day of the month falls on 
 $day_of_week = date('D', $first_day) ; 
 //Once we know what day of the week it falls on, we know how many blank days occure before it. If the first day of the week is a Sunday then it would be zero
 switch($day_of_week){ 
 case "Sun": $blank = 0; break; 
 case "Mon": $blank = 1; break; 
 case "Tue": $blank = 2; break; 
 case "Wed": $blank = 3; break; 
 case "Thu": $blank = 4; break; 
 case "Fri": $blank = 5; break; 
 case "Sat": $blank = 6; break; 
 }
 //We then determine how many days are in the current month
 $days_in_month = cal_days_in_month(0, $month, $year) ; 
 //Here we start building the table heads 
 echo "<table border=1 width=294>";
 echo "<tr><th colspan=7> $title $year </th></tr>";
 echo "<tr><td width=42>S</td><td width=42>M</td><td 
width=42>T</td><td width=42>W</td><td width=42>T</td><td 
width=42>F</td><td width=42>S</td></tr>";
 //This counts the days in the week, up to 7
 $day_count = 1;
 echo "<tr>";
 //first we take care of those blank days
 while ( $blank > 0 ) 
 { 
 echo "<td></td>"; 
 $blank = $blank-1; 
 $day_count++;
 } 
 //sets the first day of the month to 1 
 $day_num = 1;

 //count up the days, untill we've done all of them in the month
 while ( $day_num <= $days_in_month  ) 
 { 
 echo "<td> $day_num </td>"; 
 $day_num++; 
 $day_count++;

 //Make sure we start a new row every week
 if ($day_count > 7)
 {
 echo "</tr><tr>";
 $day_count = 1;
 }
 } 
 //Finaly we finish out the table with some blank details if needed
 while ( $day_count >1 && $day_count <=7 ) 
 { 
 echo "<td> </td>"; 
 $day_count++; 
 } 

 echo "</tr></table>"; 
?>

---

正如我所指出的，这部分需要有一个

mtkime（）

计算：

 //count up the days, until we've done all of them in the month
 while ( $day_num <= $days_in_month  ) { 
     echo "<td> $day_num ".date("D",mktime(0,0,0,$month,$day_num,$year))."</td>"; 
     $day_num++; 
     $day_count++;

     //Make sure we start a new row every week
     if ($day_count > 7) {
         echo "</tr><tr>";
         $day_count = 1;
     }
 } 

//计算天数，直到我们在一个月内完成所有工作
而（$day_num 7）{
回声“；
$day_count=1；
}
} 

正如我所指出的，这部分需要有一个

mtkime（）

计算：

 //count up the days, until we've done all of them in the month
 while ( $day_num <= $days_in_month  ) { 
     echo "<td> $day_num ".date("D",mktime(0,0,0,$month,$day_num,$year))."</td>"; 
     $day_num++; 
     $day_count++;

     //Make sure we start a new row every week
     if ($day_count > 7) {
         echo "</tr><tr>";
         $day_count = 1;
     }
 } 

//计算天数，直到我们在一个月内完成所有工作
而（$day_num 7）{
回声“；
$day_count=1；
}
} 

不知何故，您在某处执行

$var='Tue'

，然后执行

$var++

。无法看到代码中如何/在何处发生这种情况，但这就是为什么您有

Tue

，

Tuf

，

Tug

，等等。。。注意e->f->g。。。。你正在增加一个字符串。嗨，马克，代码现在只显示数字，我拿出了它显示的部分（星期二，星期二，星期五，拖船，等等），这是它显示数字回声“$day_num”；您必须使用填充值的数组将该值

$daysInWeek

转换为一天，该数组包含与文本值关联的数字键，如：

$daysInWeek[1]=“mon”$daysInWeek[2]=“周二”等。。。然后
echo$daysInWeek[$day_of u week]@rasclatt不，我没有。那会是什么样子？不知何故，在某个地方，你正在做
$var='Tue'
，然后
$var++
。无法看到代码中如何/在何处发生这种情况，但这就是为什么您有
Tue
，
Tuf
，
Tug
，等等。。。注意e->f->g。。。。你正在增加一个字符串。嗨，马克，代码现在只显示数字，我拿出了它显示的部分（星期二，星期二，星期五，拖船，等等），这是它显示数字回声“$day_num”；您必须使用填充值的数组将该值
$daysInWeek
转换为一天，该数组包含与文本值关联的数字键，如：
$daysInWeek[1]=“mon”$daysInWeek[2]=“周二”等。。。然后
echo$daysInWeek[$day_of u week]@rasclatt不，我没有。那会是什么样子？