INSERT issue

[TapeGun007]

I'm beating my head against the wall trying to figure out what I've changed.  This INSERT near the bottom was working.  I changed something and now it fails every time.

 

 

        if($Email == $Username && $Pass == $Password){
            {set session ID's}
 
            // $stmt->close();
            
            // If the login is successful, then go ahead and UPDATE and override all failed attempts.
            $sql = "UPDATE LoginAttempts SET LACleared = 'yes' WHERE LASalesID = '$ID'";
            if ($con->query($sql) === TRUE){
                echo "Login Successful";
            }else{
                echo "Error updating record";
            }
            // $con->close();
            
            // Redirect to crm.php
            header('Location: crm.php');
            die();
        }else{
            $sql="INSERT INTO LoginAttempts (LASalesID) VALUES ('10000')";
             
            if ($con->query($sql) === TRUE) {
                echo "New record created successfully";
            } else {
                echo "Error: " . $sql . "<br>" . $conn->error;  <---- I get this error every time.
            }
            
            $con->close();

 

So at the bottom, I get the Echo "Error: " with the sql string every single time.  I have no idea what I've done wrong.  All the other fields in the table LoginAttempts are set to default one of which is a TimeStamp in mySQL.

 

I know that the $con is fine because I use it early to verify the username is correct from another table.  What am I doing wrong here?

 

 

[Barand]

Try outputting

$con->error

instead of

$conn->error

and see if that gives a message

[TapeGun007]

Error: INSERT INTO LoginAttempts (LASalesID) VALUES ('10000')
Commands out of sync; you can't run this command now

 

Good catch Barand.  I was using code from another site and apparently didn't fix all the conn to con's.  At least I can Google this error message and see if that helps.

[TapeGun007]

I had a mySQLi prepared statement that was just before that initial IF THEN.  

 

I put $stmt->close(); just before the IF THEN and that fixed the error.

 

Thank you for helping me find the correct error message!