在PHP生成的表中显示用户的叶子
我试图创建一个PHP日历,显示每个用户及其缺勤的每月视图。这是我的桌子: 我现在看到的是(注意缺席后的空字段): 正如您所看到的,每个缺勤类型都在每个表中。例如,“Max”应该只有“Compensation”。到目前为止,我所做的是得到正确的日期。如果日期为在PHP生成的表中显示用户的叶子,php,html,mysql,Php,Html,Mysql,我试图创建一个PHP日历,显示每个用户及其缺勤的每月视图。这是我的桌子: 我现在看到的是(注意缺席后的空字段): 正如您所看到的,每个缺勤类型都在每个表中。例如,“Max”应该只有“Compensation”。到目前为止,我所做的是得到正确的日期。如果日期为2015-02-01,则日期为0(因为循环从-1开始)。现在我只使用开始日期,而不是从-到结束日期 <html> <head> <meta charset="utf-8">
2015-02-01
,则日期为0(因为循环从-1开始)。现在我只使用开始日期,而不是从-到结束日期
<html>
<head>
<meta charset="utf-8">
<link rel="stylesheet" type="text/css" href="css/style.css">
<title>Absence System</title>
</head>
<body>
<div id="container">
<?php
//mysql connection
$con = mysql_connect("localhost", "root", "");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
//select db
mysql_select_db("absence_system", $con);
$result = mysql_query("select count(1) FROM employee");
$row = mysql_fetch_array($result);
//count user
$count_user = $row[0];
//array with absences
$result2 = mysql_query("select start, end, type_FK, employee_FK FROM absences");
while ($row2 = mysql_fetch_assoc($result2)) {
$array_absences[] = $row2;
}
//count absences
$count_absences = count($array_absences);
//array with names
$result = mysql_query("select name, employee_ID FROM employee");
while ($row = mysql_fetch_assoc($result)) {
$array_user[] = $row;
}
//array with surenames
$result = mysql_query("select surename FROM employee");
while ($row2 = mysql_fetch_assoc($result)) {
$array_surname[] = $row2;
}
//find out the day from the date in a month
for ($i = 0; $i < $count_absences; $i++) {
$array_absences[$i]['start'] = substr($array_absences[$i]['start'], -2);
$array_absences[$i]['end'] = substr($array_absences[$i]['end'], -2);
$array_absences[$i]['start'] = ereg_replace("^0", "", $array_absences[$i]['start']);
$array_absences[$i]['end'] = ereg_replace("^0", "", $array_absences[$i]['end']);
$array_absences[$i]['start'] = $array_absences[$i]['start'] - 1;
}
//table with numbers
echo "<table border='1'><br />";
echo "<tr>";
for ($i = 0; $i < 32; $i++) {
if ($i == 0) {
echo "<td>", "Name", "</td>";
} else {
echo "<td>", $i, "</td>";
}
}
echo "</tr>";
echo "</table>";
//do as many times as there are users
for ($row = 0; $row < $count_user; $row++) {
echo "<table border='1'><br />";
echo "<tr>";
//table with 31 days
for ($col = 0; $col < 31; $col++) {
$true = 0;
if ($col == 0) {
//first field for the name
echo "<td>", $array_user[$row]['name'], " ", $array_surname[$row]['surename'], "</td>";
}
//absence fields
for ($i = 0; $i < $count_absences; $i++) {
if ($col == $array_absences[$i]['start']) {
echo "<td>", $array_absences[$i]['type_FK'], "</td>";
$true = 1;
}
}
//normal fields
if ($true == 0) {
echo "<td>", $col, "</td>";
}
}
echo "</tr>";
}
//table end
echo "</table>";
?>
</div>
</body>
</html>
缺勤制度
我基本上需要的是循环中的第二个条件,其中if是(而不是员工的PK,因为如果有人被删除,订单是错误的)
有什么办法吗