"将删除按钮添加到 PHP 结果表"

Question

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.

<?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>

Answer 1

You have to pass variable in delete link. You must have to pass <?php echo $contact['name']; ?><\/code> <?php echo $contact['name']; ?><\/code> name value in hidden field or pass this value in URL<\/code>

<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>

Answer 2

USe javascript

<input name="Submit2" type="button" class="button" onclick="javascript:location.href='delete.php?id=<?php echo $your_id;?>';" value="&laquo; Back" />

and in delet.php

$id=$_GET['id'];

and put $id in your sql statement.

Answer 3

You are missing to pass name in this line:

<input type="hidden" name="name" value="">

Answer 4

<input type="hidden" name="name" value="">

Answer 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");

?>

Answer 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");
?>