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Home/ Questions/Q 7816543
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Editorial Team
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Editorial Team
Asked: June 2, 20262026-06-02T05:54:11+00:00 2026-06-02T05:54:11+00:00

I want to use this page to determine if the user has either updated

  • 0

I want to use this page to determine if the user has either updated a newsletter or created a new one. It connects to the database no problem and will update, but I can not get it to insert a fresh one as it gives me this error:

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ” at line 1

If I remove the if statement and just force it to insert a new record it will work with no issues.

Any advice would be greatly appreciated, thank you.

<?php
$server = "localhost";
$username = "user";
$password = "****";
$database = "test";

$con = mysql_connect($server, $username, $password);

$title = $_POST["title"];
$body = $_POST["body"];
$transaction = "Record Added";

if (!$con)
{
    die('Could not connect: ' . mysql_error());
}

mysql_select_db($database, $con);

if(isset($_POST["id"]))
{
    $sql = "INSERT INTO newsletter (date, title, body)
    VALUES('1990-12-12', '$title', '$body')";       
}
else
{
    $id = $_POST["id"];
    $transaction = "Record Updated";
    $sql = "UPDATE newsletter SET title='".$title."', body='".$body."' WHERE newsletterId =".$id;   
}

if (!mysql_query($sql,$con))
{
    die('Error: ' . mysql_error());
}

echo $transaction;

mysql_close($con);
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-02T05:54:13+00:00Added an answer on June 2, 2026 at 5:54 am

    Your logic seems flawed, if an ID is posted you try to insert and when no ID is posted you try to update:

    if(isset($_POST["id"]))
    {
        $id = $_POST["id"];
        $transaction = "Record Updated";
        $sql = "UPDATE newsletter SET title='".$title."', body='".$body."' WHERE newsletterId =".$id;        
    }
    else
    {
        $sql = "INSERT INTO newsletter (date, title, body)
        VALUES('1990-12-12', '$title', '$body')";   
    }
    

    That is where your error comes from, your $id is empty.

    Apart from that you should look into sql injection. Switching to prepared statements is the best way to go.

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