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所以我有一个非常简单的日历的代码，但是我不知道如何查询 mysql 数据库来查找该月每一天的任何事件。我知道如何在数据库中查询该信息，但我不知道我会将它放在创建日历的代码中的什么位置，或者如何构建它或其他任何东西。我查看了许多 php 日历代码，对于每个代码我都无法从我的数据库中获取数据，我只是不知道该怎么做。任何帮助将不胜感激，谢谢。编辑日历代码以包含数据库内容：

include ("connection.php"); 

$date =time (); 
$day = date('d', $date); 
$month = date('m', $date); 
$month = $month + "02";
$year = date('Y', $date); 
$years = substr($year, 2, 2);

$first_day = mktime(0,0,0,$month, 1, $year); 
$title = date('F', $first_day); 
$day_of_week = date('D', $first_day); 
switch($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; 
}

$days_in_month = cal_days_in_month(0, $month, $year);
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>";
$day_count = 1;
echo "<tr>";

while ( $blank > 0 ) 
{ 
    echo "<td></td>"; 
    $blank = $blank-1; 
    $day_count++;
} 
$day_num = 1;
 while ( $day_num <= $days_in_month ) 
{ 

echo "<td> $day_num <br/>";


$result = mysql_query("SELECT time, length FROM hire WHERE day = '$day_num' and month =    '$month' and year = '$years'") or die ('Error: '.mysql_error ());

而 ($row = mysql_fetch_array($result)){

$time = $row['time'];
$length = $row['length'];

}

if (isset($time) and (isset($length))) {
 echo "Time: " . $time . "<br/> Length: " . $length . "<br/>";
}

"</td>";



    $day_num++; 
    $day_count++;
        if ($day_count > 7)
        {
        echo "</tr><tr>";
        $day_count = 1;
        }

} 

while ( $day_count >1 && $day_count <=7 ) 
{ 
    echo "<td> </td>"; 
    $day_count++; 
} 

echo "</tr></table>";
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1 回答 1

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您需要将数据库代码放在输出日期单元格的位置。这是代码：

<?php
include ("connection.php"); 

$date =time (); 
$day = date('d', $date); 
$month = date('m', $date); 
$month = $month + "01";
$year = date('Y', $date); 

$first_day = mktime(0,0,0,$month, 1, $year); 
$title = date('F', $first_day); 
$day_of_week = date('D', $first_day); 
switch($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; 
}

$days_in_month = cal_days_in_month(0, $month, $year);
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>";
$day_count = 1;
echo "<tr>";

while ( $blank > 0 ) 
{ 
    echo "<td></td>"; 
    $blank = $blank-1; 
    $day_count++;
} 
$day_num = 1;
 while ( $day_num <= $days_in_month ) 
{ 
    //database code here
    $result = ''; //formatted html result
    echo "<td> $day_num $result</td>"; 
    $day_num++; 
    $day_count++;
    if ($day_count > 7)
    {
        echo "</tr><tr>";
        $day_count = 1;
    }
} 

while ( $day_count >1 && $day_count <=7 ) 
{ 
    echo "<td> </td>"; 
    $day_count++; 
} 

echo "</tr></table>";
于 2013-01-20T16:06:16.797 回答