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php - 如何将mysql查询输出到html表?

转载 作者:行者123 更新时间:2023-11-29 02:27:27 25 4
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再次需要您的帮助。

所以我有一个查询,其中包含多个变量,我必须将这些变量输出到一个表中。

这里是查询:

SELECT DISTINCT
zi.zile,
ore.ora AS ore,
materii.materie
FROM
zi LEFT JOIN orar ON zi.id = orar.id_zi
LEFT JOIN ore ON ore.id = orar.id_ora
LEFT JOIN nume_scoli ON nume_scoli.id = orar.id_scoala
LEFT JOIN materii_pe_clase ON materii_pe_clase.id_scoala = nume_scoli.id
LEFT JOIN clase ON materii_pe_clase.id_clasa = clase.id AND orar.id_clasa = clase.id
LEFT JOIN elevi ON elevi.id_clasa = materii_pe_clase.id_clasa
LEFT JOIN materii ON materii.id = orar.id_materie
WHERE clase.`id`=1

ORDER BY zi.`zile`, ore.`id` ASC

这个查询的结果是这样的:

image

现在这是我在 php 中所做的:

  $oraru = "the query from up this page ";
$gaseste_oraru = mysql_query($oraru);
$numar_orar = mysql_num_rows($oraru);
if($numar_orar==0)
{
echo "Orarul nu este disponibil momentan.";
}
else
{
while($randorar=mysql_fetch_array($gaseste_oraru))
{
$ziorar = $randorar['zile'];
$oraorar = $randorar['ore'];
$materieorar = $randorar['materie'];
}
}

这是我要用数据库中的数据填充的表:

table

这里是表格的代码:

<div class="CSSTableGenerator" >
<table >
<tr>
<td>
Orar
</td>
<td>
Luni
</td>
<td >
Marti
</td>
<td>
Miercuri
</td>
<td>
Joi
</td>
<td>
Vineri
</td>
</tr>
<tr>
<td >
8:00 - 9:00
</td>
<td>
<select>
<option>Mate</option>
<option>Romana</option>
<option>Geogra</option>
</select>

</td>
<td>
<select>
<option>Mate</option>
<option>Romana</option>
<option>Geogra</option>
</select>
</td>
<td>
<select>
<option>Mate</option>
<option>Romana</option>
<option>Geogra</option>
</select>
</td>
<td>
<select>
<option>Mate</option>
<option>Romana</option>
<option>Geogra</option>
</select>
</td>

<td>
rand 1
</td>
</tr>
<tr>
<td >
9:00 - 10:00
</td>
<td>
rand 2
</td>
<td>
rand 2
</td>
<td>
rand 1
</td>
<td>
rand 1
</td>
<td>
rand 1
</td>
</tr>
<tr>
<td >
10:00 - 11:00
</td>
<td>
rand 2
</td>
<td>
rand 2
</td>
<td>
rand 1
</td>
<td>
rand 1
</td>
<td>
rand 1
</td>
</tr>
<tr>
<td >
11:00 - 12:00
</td>
<td>
rand 3
</td>
<td>
rand 3
</td>
<td>
rand 1
</td>
<td>
rand 1
</td>
<td>
rand 1
</td>
</table>
</div>

别在意下拉菜单,我不会用它们!

您是否知道如何填充它,但是如果某个时间间隔内不存在任何内容(例如 Luni 8:00-9:00),那么什么都不写在表中?

编辑

这是我尝试过的:

<div class="CSSTableGenerator" >
<table >
<tr>
<td>
Orar
</td>
<?php
{
$oraru = "the query from up this page";
$gaseste_oraru = mysql_query($oraru);

while($randorar=mysql_fetch_array($gaseste_oraru))
{
$ziorar = $randorar['zile'];
$oraorar = $randorar['ore'];
$materieorar = $randorar['materie'];
echo "<td>";
echo $randorar['zile'];
echo "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>";
echo $randorar['ore'];
echo "</td>";
echo"<td>";
echo $randorar['materie'];
echo "</td>";
echo "</tr>";
}//sfarsit while

结果如下:

bad

有什么想法吗?

最佳答案

尝试做这样的事情

$xarr=Array();
$xcols=Array();
while($randorar=mysql_fetch_array($gaseste_oraru))
{
$xcols[$randorar['zile']]++;
$xarr[$randorar['ore']][$randorar['zile']] = $randorar['materie']
}
echo "<tr><th>orar</th>";
foreach(array_keys($xcols) as $x)
echo '<th>'.$x.'</th>';
echo "</tr>\n";

foreach($xarr as $key=>$y){
echo '<tr><th>'.$key.'</th>';
foreach(array_keys($xcols) as $x)
echo '<td>'.$y[$x].'</td>';
echo "</tr>\n";
}

关于php - 如何将mysql查询输出到html表?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19067410/

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