How to delete row in MySQL table whereby PHP HTML form

Hi Everyone,
MySQL table nemed “test” is onsisted of 2 fields as shown at the screenshot hereby:

The above table contains 4 rows as shown at thescreenshot above.

My php code uploads a table that browses the above table’s rows. At the end of each row in the PHP page there is
a “submit button” to delete the current row as it shows on the following screenshot:

As you can see at the above screenshot, when I tap the “Delete” button I get an error that says that the delete
query is missing the “where” “counter” key.
I try to set the current row’s “counter” value but i’m afraid I dont know how to do so…
Here is the code:

<?php // to_forum.php
$MyHOST = 'localhost';
$MyUSER = 'user';
$MyPASS = 'pass';
$MyDB = 'test';

if (isset($_POST['delete']) && isset($_POST['MyCOUNTER']))
	$MyQUE = "DELETE FROM test WHERE counter = '$MyCOUNT'";
	if ($MyRESULT) echo "DELETE failed: $MyQUE <br>" . $MyCONNECTION->error . "<br><br>";

echo <<<_END
<form action="to_forum1.php" method="post">
Id <input type="number" name="MyID">
<input type="submit" value="ADD ID">

$MyQUE = "SELECT  counter, id FROM test ";

if (!$MyRESULT) die ("Database access failed: " . $MyCONNECTION->error);
$numOfRows = $MyRESULT->num_rows;

echo <<<_END
table, th, td {border-collapse: collapse;}
th {background-color: #f1f1c1;border: 1px solid black;}
td {border: 1px solid black;}
<table style="width:100%; border: 1px solid black;">
for ($j = 0 ; $j < $numOfRows ; $j++)
$row = $MyRESULT->fetch_array(MYSQLI_NUM);
echo <<<_END
	<td><form action="to_forum1.php" method="post">
  <input type="hidden" name="delete" value="yes">
  <input type="hidden" name="MyCOUNTER">
  <input type="submit" value="Delete"></form>
echo "</table>";

function get_post($conn, $var)
return $conn->real_escape_string($_POST[$var]);

Could anyone show me please how to set the current row’s counter value to the “Delete” query?
Thanks a lot !

The input named “MyCounter” has no value set.

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

You need to insert the relevant value to the input, eg:-

<input type="hidden" name="MyCOUNTER" value="$row['counter']">
1 Like

So I did and it worked !
Thanks a lot Sama74 !!

Glad it is working.

Now before moving on, please divert for a bit and change your code to use prepared statements:

It is not difficult and will help protect your code from sql injections. The quicker you learn how to use prepared statements the better off you will be.


Hi ahundiak,
Thanks for your advice and the link added.
I read the article to find out that where i stand right now with PHP is not a stable basis to advance to “prepared statements”. I keep your notification as well as others and once I feel I can handle it I’ll read it again and try to implement it.
Thanks a lot !

that makes switching to prepare statements even more urgent than if your PHP were written properly

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