Delete button on PHP echo table not working

Question

Below is my code for a PHP echo delete that contains a delete button. When pressed I want that entry to be deleted from the database

$result = mysql_query("SELECT * ,CONCAT (HomeScore,'-',AwayScore) AS Score, CONCAT(Against) AS Game FROM Fixture WHERE TeamID='9' ORDER BY Date DESC");
echo "<table id='customers' border='1'>;

<tr>
<th>FixtureID</th>
<th>Competition</th>
<th>Match</th>
<th>Date</th>
<th>Time</th>
<th>Score</th>
<th>test</th>
</tr>";

while($row = mysql_fetch_array($result))
{

echo "<tr>";
echo "<td>" . $row['FixtureID'] . "</td>";
echo "<td>" . $row['Competition'] . "</td>";
echo "<td>" . $row["Against"] . "</td>";
echo "<td>" . $row['Date'] . "</td>";
echo "<td>" . $row['Time'] . "</td>";
echo "<td>" . $row['Score'] . "</td>";
echo "<td><form method=post>
                    <input name=id type=hidden value='".$row['FixtureID']."';>
                    <input type=submit name=submit value=Delete>
                    </form></td>";
                    echo "</tr>";
echo "</tr>";  
}
}
echo "</table>";

// delete record
if($_SERVER['REQUEST_METHOD'] == "POST")
{
if(isset($_POST['FixtureID']))
{
$id = FixtureID;
$sql = mysql_query("DELETE FROM Fixture WHERE FixtureID =$id");
if(!$sql)
{
    echo ("Could not delete rows" .mysql_error());
}
}
}

How do I get this to work? Also FixtureID is stored as an integer in the database.

Your looking at the variable `$_POST['FixtureID']`, shouldn't it be `$_POST['id']`? Might be worth that putting the `//delete record` block before the display code, that way you don't need to refresh the page once you have clicked the delete button. — Dom, Apr 15 '14 at 12:15
Use action in the form where you want to post your data and rather creating form in each row you can use anchor link. pass id in query string or using ajax. — Trimantra Software Solution, Apr 15 '14 at 12:16
You are also creating form in a loop which means you are creating a button with a same name "id". this results in to array of the elements of same name. — Trimantra Software Solution, Apr 15 '14 at 12:18
**Danger**: You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). You are also **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that a modern API would make it easier to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, Apr 15 '14 at 12:20
@TrimantraSoftwareSolution — Don't use a link. Links make GET requests. Deleting things should be done with a POST request (or a DELETE request in a RESTful API). — Quentin, Apr 15 '14 at 12:21
@TrimantraSoftwareSolution — For the purposes of submitting a form, only form controls in that form count. So only the input that is in the current form matters. There is no array. — Quentin, Apr 15 '14 at 12:21

David · Accepted Answer · 2014-04-15T13:42:08.723

Try with these changes :

// delete record
if(isset($_POST['submit']) && isset($_POST['id']) && !empty($_POST['id'])) {
  $id = $_POST['id'];
  $sql = mysql_query("DELETE FROM Fixture WHERE FixtureID =$id");
  if(!$sql) {
    echo ("Could not delete rows" .mysql_error());
  }
}

Edit :

// Delete record if ID submitted
if(isset($_POST['submit']) && isset($_POST['id']) && !empty($_POST['id'])) {
  $id = $_POST['id'];
  $sql = mysql_query("DELETE FROM Fixture WHERE FixtureID =$id");
  if(!$sql) {
    echo ("Could not delete rows" .mysql_error());
  }
}

// Get datas from BDD
$result = mysql_query("SELECT * ,CONCAT (HomeScore,'-',AwayScore) AS Score, CONCAT(Against) AS Game FROM Fixture WHERE TeamID='9' ORDER BY Date DESC");

// Display data
echo "
<table id='customers' border='1'>
<tr>
<th>FixtureID</th>
<th>Competition</th>
<th>Match</th>
<th>Date</th>
<th>Time</th>
<th>Score</th>
<th>test</th>
</tr>";

// For each result
while($row = mysql_fetch_array($result)) {

    echo "<tr>";
    echo "<td>" . $row['FixtureID'] . "</td>";
    echo "<td>" . $row['Competition'] . "</td>";
    echo "<td>" . $row["Against"] . "</td>";
    echo "<td>" . $row['Date'] . "</td>";
    echo "<td>" . $row['Time'] . "</td>";
    echo "<td>" . $row['Score'] . "</td>";
    echo "<td><form method=post>
                    <input name=id type=hidden value='".$row['FixtureID']."';>
                    <input type=submit name=submit value=Delete>
                    </form></td>";
                    echo "</tr>";
    echo "</tr>";
}
echo "</table>";

@Aliasse Thank you cant accept the answer yet as not enough time has gone by but i will. How do i get it to refresh after the click so it disappears automatically? — user3535330, Apr 15 '14 at 12:24
@user3535330 — Move the code that deletes the row so it runs before the code that lists all the rows. — Quentin, Apr 15 '14 at 12:27
@Quentin is right. Move this code before your data table and before your data query. — David, Apr 15 '14 at 12:36
@Aliassse and I've updated my code to show everything sorry where exactly does it go im having no luck — user3535330, Apr 15 '14 at 12:43
@Quentin I've updated my code to show everything sorry where exactly does it go im having no luck — user3535330, Apr 15 '14 at 12:44
@user3535330 Put your delete code before your query. This way, you'll delete before you get datas. — David, Apr 15 '14 at 13:36

score 0 · Answer 2 · edited Apr 15 '14 at 12:58

0

$id value is not properly set

$id = $_POST['FixtureID'];

edited Apr 15 '14 at 12:58

Jeff B

8,572
17
61
140

answered Apr 15 '14 at 12:18

Om Prakash

54
1
1
5

score 0 · Answer 3 · answered Apr 15 '14 at 12:20

0

just try this

if(isset($_POST['submit']))
{

$id = $_POST['id']; //  here you should use form post value
$sql = mysql_query("DELETE FROM Fixture WHERE FixtureID =$id");
if(!$sql)
{
    echo ("Could not delete rows" .mysql_error());
}
}
}

answered Apr 15 '14 at 12:20

Dinesh

4,066
5
21
35

Delete button on PHP echo table not working

3 Answers3