我想用 PHP 和 MySQL 做一个缺勤系统。每个缺勤都有一个休假类型。我想在生成的 HTML 表格中显示这种类型(例如生病)。左侧第一个字段显示用户名，其他为 31 个字段(如月份)。
我可以在表格中输入类型，但是我不知道如何只在正确的表格中显示它们。例如，“Max”的 PK 0 和缺席类型为“Emergency”和“ASDF”，PK 2 和“Compensation”，但类型在两个表中。
Array (
[0] =>
Array ( [start] => 0 [end] => 27 [type_FK] => Compensation [employee_FK] => 0 )
[1] =>
Array ( [start] => 1 [end] => 3 [type_FK] => Emergency [employee_FK] => 2 )
)
如何在该表中只显示正确用户的休假类型？
<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
$con = mysql_connect("localhost", "root", "");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("absence_system", $con);
$result = mysql_query("select count(1) FROM employee");
$row = mysql_fetch_array($result);
$count_user = $row[0];
$result2 = mysql_query("select start, end, type_FK, employee_FK FROM absences");
while ($row2 = mysql_fetch_assoc($result2)) {
$array_absences[] = $row2;
}
$count_absences = count($array_absences);
$result = mysql_query("select name FROM employee");
while ($row = mysql_fetch_assoc($result)) {
$array_user[] = $row;
}
$result = mysql_query("select surename FROM employee");
while ($row2 = mysql_fetch_assoc($result)) {
$new_array2[] = $row2;
}
for ($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;
echo $array_absences[$i]['start'], "<br>";
}
print_r($array_absences);
echo "<table border='1'><br />";
echo "<tr>";
for ($i = 0; $i < 32; $i++) {
if ($i == 0) {
echo "<td>", "Name", "</td>";
} else {
echo "<td>", $i, "</td>";
}
}
echo "</tr>";
echo "</table>";
for ($row = 0; $row < $count_user; $row++) {
echo "<table border='1'><br />";
echo "<tr>";
//Tabelle mit 31 Tagen generieren
for ($col = 0; $col < 32; $col++) {
$true = 0;
if ($col == 0) {
//Name in die ersten Felder schreiben
echo "<td>", $array_user[$col]['name'], " ", $new_array2[$col]['surename'], "</td>";
}
for ($i = 0; $i < $count_absences; $i++) {
if ($col == $array_absences[$i]['start']) {
echo "<td>", $array_absences[$i]['type_FK'], "</td>";
$true = 1;
}
}
//Normale Felder
if ($true == 0) {
echo "<td>", $col, "</td>";
}
}
echo "</tr>";
}
echo "</table>";
?>
</div>
</body>
</html>
-- phpMyAdmin SQL Dump
-- version 4.2.11
-- http://www.phpmyadmin.net
--
-- Host: 127.0.0.1
-- Erstellungszeit: 13. Feb 2015 um 16:07
-- Server Version: 5.6.21
-- PHP-Version: 5.5.19
SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";
/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;
--
-- Datenbank: `absence_system`
--
-- --------------------------------------------------------
--
-- Tabellenstruktur für Tabelle `absences`
--
CREATE TABLE IF NOT EXISTS `absences` (
`absences_ID` int(11) NOT NULL,
`employee_FK` int(11) NOT NULL,
`start` date NOT NULL,
`end` date NOT NULL,
`approved` tinyint(1) NOT NULL,
`comment` varchar(45) NOT NULL,
`type_FK` varchar(50) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- --------------------------------------------------------
--
-- Tabellenstruktur für Tabelle `employee`
--
CREATE TABLE IF NOT EXISTS `employee` (
`employee_ID` int(11) NOT NULL,
`name` varchar(45) NOT NULL,
`surename` varchar(45) NOT NULL,
`on_offshore_FK` int(11) NOT NULL,
`location_FK` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- --------------------------------------------------------
--
-- Tabellenstruktur für Tabelle `location`
--
CREATE TABLE IF NOT EXISTS `location` (
`location_ID` int(7) NOT NULL,
`country` varchar(45) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- --------------------------------------------------------
--
-- Tabellenstruktur für Tabelle `on_offshore`
--
CREATE TABLE IF NOT EXISTS `on_offshore` (
`on_offshore_ID` int(11) NOT NULL,
`on_off` varchar(45) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- --------------------------------------------------------
--
-- Tabellenstruktur für Tabelle `type`
--
CREATE TABLE IF NOT EXISTS `type` (
`type` varchar(50) NOT NULL DEFAULT ''
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
--
-- Indizes der exportierten Tabellen
--
--
-- Indizes für die Tabelle `absences`
--
ALTER TABLE `absences`
ADD PRIMARY KEY (`absences_ID`), ADD KEY `employee_FK` (`employee_FK`), ADD KEY `type_FK` (`type_FK`);
--
-- Indizes für die Tabelle `employee`
--
ALTER TABLE `employee`
ADD PRIMARY KEY (`employee_ID`), ADD KEY `on_offshore_FK` (`on_offshore_FK`), ADD KEY `location_FK` (`location_FK`);
--
-- Indizes für die Tabelle `location`
--
ALTER TABLE `location`
ADD PRIMARY KEY (`location_ID`);
--
-- Indizes für die Tabelle `on_offshore`
--
ALTER TABLE `on_offshore`
ADD PRIMARY KEY (`on_offshore_ID`);
--
-- Indizes für die Tabelle `type`
--
ALTER TABLE `type`
ADD PRIMARY KEY (`type`);
--
-- Constraints der exportierten Tabellen
--
--
-- Constraints der Tabelle `absences`
--
ALTER TABLE `absences`
ADD CONSTRAINT `absences_ibfk_2` FOREIGN KEY (`employee_FK`) REFERENCES `employee` (`employee_ID`),
ADD CONSTRAINT `absences_ibfk_3` FOREIGN KEY (`type_FK`) REFERENCES `type` (`type`);
--
-- Constraints der Tabelle `employee`
--
ALTER TABLE `employee`
ADD CONSTRAINT `employee_ibfk_1` FOREIGN KEY (`on_offshore_FK`) REFERENCES `on_offshore` (`on_offshore_ID`),
ADD CONSTRAINT `employee_ibfk_2` FOREIGN KEY (`location_FK`) REFERENCES `location` (`location_ID`);
/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;
最佳答案
按照这些思路尝试一些事情:
获取一个月的缺勤列表:Select * from abnissions_table。
获取员工列表:从 abnances_table 中选择 *。
创建一个空的 $employees 数组。循环遍历缺席结果，并针对每次缺席运行如下内容:
for($i=0;$i<count($absences_array);$i++){
if(!$employees[$absences_array[$i]['employee_FK']]['dates']){
$employees[$absences_array[$i]['employee_FK']]['dates'] = array();
}
$employees[$absences_array[$i]['employee_FK']]['dates'][$absences_array[$i]['employee_FK']['start']];
for( $j=$absences_array[$j]['employee_FK']['start'];$j<$absences_array[$j]['employee_FK']['end']; $j++){
$employees[$absences_array[$i]['employee_FK']]['dates'][$j];
}
}
现在你有一个字典的字典 - 这意味着像 $employees[$employee_id]['dates'][$date] 这样的东西会给你缺席类型或 null。
关于php - 在 PHP 生成的表中显示 MySQL 用户数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28502613/