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问题:
I have outputted the results of a MySQL table to an HTML table. In the last column, I want to add a delete option which calls another form and deletes the user. I can't seem to get it to work though.
This is my code for the results page:
<?php
$contacts = mysql_query("
SELECT * FROM contacts ORDER BY ID ASC") or die( mysql_error() );
// If results
if( mysql_num_rows( $contacts ) > 0 )
?>
<table id="contact-list">
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Telephone</th>
<th>Address</th>
<th>Delete</th>
</tr>
</thead>
<tbody>
<?php while( $contact = mysql_fetch_array( $contacts ) ) : ?>
<tr>
<td class="contact-name"><?php echo $contact['name']; ?></td>
<td class="contact-email"><?php echo $contact['email']; ?></td>
<td class="contact-telephone"><?php echo $contact['telephone']; ?></td>
<td class="contact-address"><?php echo $contact['address']; ?></td>
<td class="contact-delete"><form action='delete.php' method="post">
<input type="hidden" name="name" value="">
<input type="submit" name="submit" value="Delete">
</form></td>
</tr>
<?php endwhile; ?>
</tbody>
</table>
and, this is my delete.php script
<?php
//Define the query
$query = "DELETE FROM contacts WHERE name={$_POST['name']} LIMIT 1";
//sends the query to delete the entry
mysql_query ($query);
if (mysql_affected_rows() == 1) {
//if it updated
?>
<strong>Contact Has Been Deleted</strong><br /><br />
<?php
} else {
//if it failed
?>
<strong>Deletion Failed</strong><br /><br />
<?php
}
?>
Pretty sure I'm just missing something, but I can't figure out what that is :(
回答1:
You have to pass variable in delete link. You must have to pass <?php echo $contact['name']; ?>
name value in hidden field or pass this value in URL
Replace
<td class="contact-delete">
<form action='delete.php' method="post">
<input type="hidden" name="name" value="">
<input type="submit" name="submit" value="Delete">
</form>
</td>
With
<td class="contact-delete">
<form action='delete.php?name="<?php echo $contact['name']; ?>"' method="post">
<input type="hidden" name="name" value="<?php echo $contact['name']; ?>">
<input type="submit" name="submit" value="Delete">
</form>
</td>
回答2:
USe javascript
<input name="Submit2" type="button" class="button" onclick="javascript:location.href='delete.php?id=<?php echo $your_id;?>';" value="« Back" />
and in delet.php
$id=$_GET['id'];
and put $id in your sql statement.
回答3:
<input type="hidden" name="name" value="">
You are missing a value which wil be picked up by this line in your delete file.
$query = "DELETE FROM contacts WHERE name={$_POST['name']} LIMIT 1";
Right now it isn't receiving anything, which is why it will not work.
So add a value to it and it will work. Example:
<input type="hidden" name="name" value="<?php echo $contact['name']; ?>">
回答4:
You are missing to pass name in this line:
<input type="hidden" name="name" value="">
You need to have something (<?php echo $contact['name']; ?>
) in the value
attribute.
BTW, do not use deprecated mysql_*
functions, use PDO or mysqli_*
instead.
回答5:
At first, you cannot write the code in that way, the code has no protection against SQL injection.
1) Try to use primary ID's instead of using a name (what happens if 2 people has the same name?).
So, you can create a hidden field to know what person you are handling with.
<input type="hidden" name="contact_id" value="<?php $contact['contact_id']; ?>">
2) Sanitize variables to avoid attacks:
<?php $contact_id = isset($_POST['contact_id'])?intval($_POST['contact_id']):0;
// proceed with the query
if($contact_id>0) { $query = "DELETE FROM contacts WHERE contact_id = '$contact_id'";
}
// redirect to the main table with header("location: main.php");
?>
回答6:
This is a php delete process script you link this code in your delete button
<?php
$link=mysql_connect("localhost","root","");
mysql_select_db("photo",$link);
if(isset($_GET["id"])
$photoid=$_GET['id'];
$sql="delete from photos where photoid=".$_GET['id'];
$result=mysql_query($sql,$link);
header("location:home.php?ok");
?>