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Home/ Questions/Q 6012553
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T02:25:15+00:00 2026-05-23T02:25:15+00:00

I have strange problem on PHP page, I am trying to insert new row

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I have strange problem on PHP page, I am trying to insert new row to table users (SQLite3) like this

 $id=$db->insert_user("'".$parameters['first_name']."'", "'".$parameters['last_name']."'", "'".$parameters['email']."'", "'".$parameters['username']."'", "'".$parameters['password']."'", "'".$parameters['privilege']."'");

where

     try {
            $db = new SQLite3($this->db_path);

           echo "INSERT INTO users(first_name,last_name,email,username,password,privilege) VALUES('$first_name','$last_name','$email','$username','$password','$privilege');";
           $db->query("INSERT INTO users(first_name,last_name,email,username,password,privilege) VALUES('$first_name','$last_name','$email','$username','$password','$privilege');");

            $last_id = $db->lastInsertRowID();
            echo ''.$last_id;
            $db->close();
            return $last_id;
        } catch (Exception $e) {
            echo 'Exception caught: ', $e->getMessage(), "\n";
        }

INSERT INTO users(first_name,last_name,email,username,password,privilege) VALUES(''drkosh'',''drkosh'',''drkosh'',''drkosh'',''drkosh'',''ignore'');

result from echo is

INSERT INTO users(first_name,last_name,email,username,password,privilege) VALUES(''drkosh'',''drkosh'',''drkosh'',''drkosh'',''drkosh'',''ignore'');

Error message is strange, it says in line 35, but 35 is in switch and I never pass through 35, have breaks. Does anybody have some clue what is wrong ? (all columns are ok and table name)

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T02:25:16+00:00Added an answer on May 23, 2026 at 2:25 am

    look at the sql query, there is ''drkosh'', but there should only be 'drkosh'

    INSERT INTO users(first_name,last_name,email,username,password,privilege) VALUES(''drkosh'',''drkosh'',''drkosh'',''drkosh'',''drkosh'',''ignore'');
    

    try this:

    $id=$db->insert_user($parameters['first_name'], $parameters['last_name'], $parameters['email'], $parameters['username'], $parameters['password'], $parameters['privilege']);
    
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