每个值都显示在一个新的行 HTML 表中 [英] Each value is displayed in a new row HTML table
问题描述
我目前在一个小型网站上工作，我的同事可以在其中查看他们的工作时间.
I'm currently working on a small website where my colleagues can see their work times.
- 我的公鸡"桌子是这样的:
- rooster_id int(2) Auto_increment 主键.
- personeel_id int(2)//personeel 在荷兰语中表示工作人员(我正在使用内连接来获取此人的姓名).
- dag varchar(10)//dag 表示一天.
- start varchar(5)//start 是员工必须开始的时间工作.
- eind varchar(5)//eind 表示结束.
- 基准日期
- weeknummer int(2)//weeknummer 表示周数.
显示:
---------------------------------------------------------------------
|rooster_id|personeel_id|dag |start|eind |datum |weeknummer|
|1 |1 |Tuesday |12:00|21:00|2016-05-10|19 |
|1 |1 |Wednesday|15:00|21:00|2016-05-11|19 |
|1 |2 |Monday |08:00|18:30|2016-05-10|19 |
---------------------------------------------------------------------
我在 HTML 标签上方的 php 代码:
My php code above HTML tag:
include_once 'config.php';
$people = "SELECT DISTINCT * FROM personeel INNER JOIN rooster ON rooster.personeel_id = personeel.id";
$result = mysql_query($people);
我在table标签内的php代码:
My php code inside the table tag:
<table cellpadding="1" width="100%" cellspacing="1" class="box-inhoud">
<?php
while($row = mysql_fetch_array($result)){
echo "<tr>";
echo "<td>".$row['naam']."</td>";
if($row['dag'] === "maandag"){
echo "<td class='td-midden'>".$row['start']." - ".$row['eind']."</td>";
} else {
echo "<td class='td-midden'>Vrij</td>";
}
if($row['dag'] === "dinsdag"){
echo "<td class='td-midden'>".$row['start']." - ".$row['eind']."</td>";
} else {
echo "<td class='td-midden'>Vrij</td>";
}
if($row['dag'] === "woensdag"){
echo "<td class='td-midden'>".$row['start']." - ".$row['eind']."</td>";
} else {
echo "<td class='td-midden'>Vrij</td>";
}
if($row['dag'] === "donderdag"){
echo "<td class='td-midden'>".$row['start']." - ".$row['eind']."</td>";
} else {
echo "<td class='td-midden'>Vrij</td>";
}
if($row['dag'] === "vrijdag"){
echo "<td class='td-midden'>".$row['start']." - ".$row['eind']."</td>";
} else {
echo "<td class='td-midden'>Vrij</td>";
}
if($row['dag'] === "zaterdag"){
echo "<td class='td-midden'>".$row['start']." - ".$row['eind']."</td>";
} else {
echo "<td class='td-midden'>Vrij</td>";
}
if($row['dag'] === "zondag"){
if($row['start'] === "00:00:00"){
echo "<td class='td-midden'>Feestdag</td>";
} else {
echo "<td class='td-midden'>".$row['start']." - ".$row['eind']."</td>";
}
} else {
echo "<td class='td-midden'>Vrij</td>";
}
echo "</tr>";
}
?>
</table>
这会显示这样的数据:
------------------------------------------------------
|Employee's|Monday |Tuesday |Wednesday |
|Tom |Vrij |12:00 - 21:00|Vrij |
|Tom |Vrij |Vrij |15:00 - 21:00|
|Jack |08:00 - 18:30|Vrij |Vrij |
------------------------------------------------------
它还显示星期四、星期五、星期六、星期日，但我不想在 stackoverflow 上写下所有的日子，因为这些日子的值是vrij".
It also displays Thursday, Friday, Saturday, Sunday but I didn't want to write all of the days here on stackoverflow since those will have "vrij" as value.
但问题是，我想要这样:
But the thing is, I want to have it like this:
------------------------------------------------------
|Employee's|Monday |Tuesday |Wednesday |
|Tom |Vrij |12:00 - 21:00|15:00 - 21:00|
|Jack |08:00 - 18:30|Vrij |Vrij |
------------------------------------------------------
所以每个员工只有 1 行，其中列出了他们的所有时间.
So every employee does only have 1 row with all their times listed in it.
我是这个网站的新手，所以我希望我解释得很好，所以你们可以轻松地帮助我.
I'm new on this website so I hope I explained everything good, so you guys can easily help me.
是的，我知道，我使用 MySQL 是因为我想保持简单，而且我从未使用过 MySQLi 或 PDO.
And yes I know, I'm using MySQL because I want to keep it simple and I have never worked with MySQLi nor PDO.
推荐答案
这里是整个示例:
<?php
$con = mysqli_connect('localhost', 'root', '', 'test') or die(mysqli_error($con));
$query = "SELECT * FROM personeel INNER JOIN rooster ON rooster.personeel_id = personeel.id";
$result = mysqli_query($con, $query) or die(mysqli_error($con));
if (mysqli_num_rows($result) > 0) {
$arr = array();
$nam = array();
$day = array('Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday');
while ($row = mysqli_fetch_assoc($result)) {
$nam[$row['id']] = $row['naam'];
$arr[$row['id']][$row['dag']] = $row['start'] . ' - ' . $row['eind'];
}
echo '<table border="1">';
echo '<tr>';
echo '<td>Employees</td>';
foreach($day as $d){
echo '<td>'.$d.'</td>';
}
echo '</tr>';
foreach ($nam as $k=>$n){
echo '<tr>';
echo '<td>'.$n.'</td>';
foreach ($day as $d){
if(isset($arr[$k][$d])){
echo '<td>'.$arr[$k][$d].'</td>';
}else{
echo '<td>Virj</td>';
}
}
echo '</tr>';
}
echo '</table>';
}
?>
输出为:
一些细节:
$nam 数组存储了从 personeel 中获取的所有姓名以及他的 id 关系.
$nam array stores all fetched names from personeel together with his id relation.
Array
(
[1] => Tom
[2] => Jack
)
$arr 是包含人员 ID 和他的数据之间关系的主数组:
$arr is main array which contains relation between personeel id and his data:
Array
(
[1] => Array
(
[Tuesday] => 12:00 - 21:00
[Wednesday] => 15:00 - 21:00
)
[2] => Array
(
[Monday] => 08:00 - 18:30
)
)
$day 是包含工作日的静态数组.
$day is static array with week days.
遍历 $nam 和 $arr 数组，即可获得所需的表，没有任何重复项.另外提供的解决方案使用了 mysqli_* 函数，所以你可以看到如何使用它们.
Walking trough both $nam and $arr arrays gives you desired table without any duplicates. In addition provided solution uses mysqli_* functions, so you can see how to use them.
希望这能帮助您解决问题.:)
Hope this will help you to solve your problem. :)
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