我的php有问题,生成一个表,从SQL数据库请求数据,并将数据存储在表中。添加和访问PHP生成的表中的按钮ID
表格中每一行的第一个单元格包含一个下拉按钮,该按钮链接到删除该行的delete.php脚本。它还链接到一个用于修改行的单元格的modif.php脚本。 我的问题是,我需要访问带有ID的下拉按钮,以知道要删除哪一行。
我真的不知道如何将我的下拉按钮与ID链接,并在我的脚本中访问它们。
下面的代码:
<?php
$con=mysqli_connect("localhost","root","icare","icare1");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM magasin");
echo "<table border='1'>
<tr>
<th>code</th>
<th>ip</th>
<th>ads</th>
<th>region</th>
<th>adress</th>
<th>name</th>
<th>email</th>
<th>number</th>
<th>gtc</th>
<th>date</th>
</tr>";
$indexB = array();
$i = 0;
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>
<div class='dropdown'>
<button id=$indexB[$i] class='dropbtn'>▶</button>
<div class='dropdown-content'>
<a href='modif.php'>Modifier</a>
<a href='delete.php'>Supprimer</a>
</div>
".$row['code']."
</div>
</td>";
echo "<td><div>" . $row['ip'] . "</div></td>";
echo "<td><div>" . $row['ads'] . "</div></td>";
echo "<td><div>" . $row['region'] . "</div></td>";
echo "<td><div>" . $row['adress'] . "</div></td>";
echo "<td><div>" . $row['name'] . "</div></td>";
echo "<td><div>" . $row['email'] . "</div></td>";
echo "<td><div>" . $row['number'] . "</div></td>";
echo "<td><div>" . $row['gtc'] . "</div></td>";
echo "<td><div>" . $row['date'] . "</td>";
echo "</tr>";
$i++;
}
echo "</table>";
mysqli_close($con);
?>
这里是delete.php:
<?php
$connection = mysqli_connect("localhost", "root", "icare", "icare1");
if($connection === false){
die("Connection failed " . mysqli_connect_error());
};
//$id =
$sql = "DELETE FROM magasin WHERE Code=".$id;
//$result = mysqli_query($connection,$sql);
if(mysqli_query($connection, $sql)){
echo "Done !";
} else{
echo "Failed : $sql. " . mysqli_error($connection);
}
mysqli_close($connection);
?>
我开始一个indexB []存储的下拉菜单标识,但我不知道我”米做对了。 最后我想将我的按钮链接到代码属性,然后使用代码属性删除与我的SQL查询行。
我是新来的,所以...对不起,如果我做了或要求一些简单的愚蠢。
UPDATE: 要mikrafizik:
我想你的答案,但它不工作。我只得到“1”> Supprimer”这seemsi有在href的一个问题,但我只是无法找到为什么 我不知道是什么我忘了,所以如果你看到一些错误:。
<?php
$con=mysqli_connect("localhost","root","icare","icare1");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM magasin");
echo "<table border='1'>
<tr>
<th>Code</th>
<th>Adresse IP</th>
<th>Adresse ADS</th>
<th>Région</th>
<th>Adresse</th>
<th>Nom du directeur</th>
<th>Mail</th>
<th>Téléphone</th>
<th>GTC</th>
<th>Date d'installation</th>
</tr>";
$data = mysqli_fetch_array($result);
?>
<table>
<?php foreach ($data as $key => $row):?>
<tr>
<td>
<div class='dropdown-content'>
<button class='dropbtn'>▶</button>
<!-- <a href="modif.php?id=<?=$row['id']?>">Modifier</a> -->
<a href="delete.php?id=<?php echo $row['id']?>">Supprimer</a>
</div>
</td>
<td><div><?php echo $row['AdresseIP'];?></div></td>
<td><div><?php echo $row['AdresseADS'];?></div></td>
<td><div><?php echo $row['Region'];?></div></td>
<td><div><?php echo $row['Adresse'];?></div></td>
<td><div><?php echo $row['NomDirecteur'];?></div></td>
<td><div><?php echo $row['Mail'];?></div></td>
<td><div><?php echo $row['Tel'];?></div></td>
<td><div><?php echo $row['Gtc'];?></div></td>
<td><div><?php echo $row['DateInstall'];?></td>
</tr>
<?php endforeach; ?>
</table>
<?mysqli_close($con);?>
delete.php:
<?php
$connection = mysqli_connect("localhost", "root", "icare", "icare1");
if($connection === false){
die("Connexion échouée " . mysqli_connect_error());
};
$id = $_GET['id'];
$sql = "DELETE FROM magasin WHERE Code=".$id;
$result = mysqli_query($connection,$sql);
if($result){
echo "Enregistrement réussi !";
} else{
echo "Enregistrement échoué : $sql. " . mysqli_error($connection);
}
mysqli_close($connection);
?>
只需添加唯一标识符链接的URL从该行删除。通过链接传递参数。所以基本上这个网址会看起来像delete.php?id = 1 –
谢谢,我读了关于在URL中传递参数,但不知道如何去做。 –