我的php有問題,生成一個表,從SQL數據庫請求數據,並將數據存儲在表中。添加和訪問PHP生成的表中的按鈕ID
表格中每一行的第一個單元格包含一個下拉按鈕,該按鈕鏈接到刪除該行的delete.php腳本。它還鏈接到一個用於修改行的單元格的modif.php腳本。 我的問題是,我需要訪問帶有ID的下拉按鈕,以知道要刪除哪一行。
我真的不知道如何將我的下拉按鈕與ID鏈接,並在我的腳本中訪問它們。
下面的代碼:
<?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);
?>
這裏是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);
?>
我開始一個indexB []存儲的下拉菜單標識,但我不知道我」米做對了。 最後我想將我的按鈕鏈接到代碼屬性,然後使用代碼屬性刪除與我的SQL查詢行。
我是新來的,所以...對不起,如果我做了或要求一些簡單的愚蠢。
UPDATE: 要mikrafizik:
我想你的答案,但它不工作。我只得到「1」> Supprimer」這seemsi有在href的一個問題,但我只是無法找到爲什麼 我不知道是什麼我忘了,所以如果你看到一些錯誤:。
<?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);
?>
只需添加唯一標識符鏈接的URL從該行刪除。通過鏈接傳遞參數。所以基本上這個網址會看起來像delete.php?id = 1 –
謝謝,我讀了關於在URL中傳遞參數,但不知道如何去做。 –