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What is the difference between MySQL, MySQLi and PDO? [closed]

What is the difference between MySQL, MySQLi and PDO?

Which one is the best suited to use with PHP-MySQL?

Which is fastest in PHP- MySQL or MySQLi?

I'd like to know if anyone has any first-hand experience with this dichotomy. A few blogs say the mysql extension is faster than mysqli. Is this true?

And I'm only asking about speed. I know mysqli has features that are not present in the older extension.

PHP and MySQLi close()

I am new to MySQL and PHP and am attempting to make my own CMS to help make managing my websites easier. Can someone explain mysqli's close() function?

1. Is it necessary?
2. What exactly does it do?
3. I heard that after PHP runs its script that it closes the connection, is that true?
4. Lastly, is there a security issue when not closing your connection to the database?

Fatal error: Class 'MySQLi' not found

I am doing a tutorial and am getting this error:

Fatal error: Class 'MySQLi' not found (LONG URL) on line 8

The code on line 8 is:

$mysqli = new MySQLi($db_server, $db_user, $db_pass, $db_name) or die(mysqli_error());

I saw online someone said to see if it was turned on in my phpinfo(), but there wasn't anything listed in there under for "mysqli".

Also, I am running PHP version 5.2.5

Why shouldn't I use mysql_* functions in PHP?

What are the technical reasons why I shouldn't use mysql_* functions? (e.g. mysql_query(), mysql_connect() or mysql_real_escape_string())?

Why should I use something else even if they work on my site?

Example of how to use bind_result vs get_result

I would like to see an example of how to call using bind_result vs. get_result and what would be the purpose of using one over the other.

Also the pro and cons of using each.

What is the limitation of using either and is there a difference.

How to echo a MySQLi prepared statement?

I'm playing around with MySQLi at the moment, trying to figure out how it all works. In my current projects I always like to echo out a query string while coding, just to make sure that everything is correct, and to quickly debug my code. But... how can I do this with a prepared MySQLi statement?

Example:

$id = 1;
$baz = 'something';

if ($stmt = $mysqli->prepare("SELECT foo FROM bar WHERE id=? AND baz=?")) {
  $stmt->bind_param('is',$id,$baz);
  // how to preview this prepared query before acutally executing it?
  // $stmt->execute();
}

I've been going through this list (http://www.php.net/mysqli) but without any luck.

EDIT

Well, if it's not possible from within MySQLi, maybe I'll stick with something like this:

function preparedQuery($sql,$params) {
  for ($i=0; $i<count($params); $i++) {
    $sql = preg_replace('/\?/',$params[$i],$sql,1);
  }
  return $sql;
}

$id = 1;
$baz = 'something';

$sql = "SELECT foo FROM bar WHERE id=? AND baz=?";

echo preparedQuery($sql,array($id,$baz));

// outputs: SELECT foo FROM bar WHERE id=1 AND baz=something

Far from perfect obviously, since it's still pretty redundant — something I wanted to prevent — and it also doesn't give me an idea as to what's being done with the data by MySQLi. But I guess this way I can quickly see if all the data is present and in the right place, and it'll save me some time compared to fitting in the variables manually into the query — that can be a pain with many vars.

Pass by reference problem with PHP 5.3.1

Ok, this is a weird problem, so please bear with me as I explain.

We upgraded our dev servers from PHP 5.2.5 to 5.3.1.

Loading up our code after the switch, we start getting errors like:

Warning: Parameter 2 to mysqli_stmt::bind_param() expected to be a reference, value given in /home/spot/trunk/system/core/Database.class.php on line 105

the line mentioned (105) is as follows:

call_user_func_array(Array($stmt, 'bind_param'), $passArray);

we changed the line to the following:

call_user_func_array(Array($stmt, 'bind_param'), &$passArray);

at this point (because allow_call_time_pass_reference) is turned off, php throws this:

Deprecated: Call-time pass-by-reference has been deprecated in /home/spot/trunk/system/core/Database.class.php on line 105

After trying to fix this for some time, I broke down and set allow_call_time_pass_reference to on.

That got rid of the Deprecated warning, but now the Warning: Parameter 2 to mysqli_stmt::bind_param() expected to be a reference warning is throwing every time, with or without the referencing.

I have zero clue how to fix this. If the target method was my own, I would just reference the incoming vars in the func declaration, but it's a (relatively) native method (mysqli).

Has anyone experienced this? How can I get around it?

Thank you.

Headers and client library minor version mismatch

In PHP I'm getting the following warning whenever I try to connect to a database (via mysql_connect)

Warning: mysql_connect(): Headers and client library minor version mismatch. Headers:50162 Library:50524

In my php -i output I have the following values listed under mysqli

Client API library version => 5.5.24

Client API header version => 5.1.62

I've tried updating php5-mysql and php but I'm already at the latest version of both of them. How do I go about updating the header version so I stop seeing this warning?

EDIT

My MySQL files should all be updated to be the latest version:

$ apt-get install mysql.*5.5
. . .
mysql-client-5.5 is already the newest version.
mysql-server-core-5.5 is already the newest version.
mysql-server-5.5 is already the newest version.
mysql-testsuite-5.5 is already the newest version.
mysql-source-5.5 is already the newest version.

Removing old versions

$ apt-get remove mysql.*5.1
. . .
Package handlersocket-mysql-5.1 is not installed, so not removed
Package mysql-cluster-client-5.1 is not installed, so not removed
Package mysql-cluster-server-5.1 is not installed, so not removed
Package mysql-client-5.1 is not installed, so not removed
Package mysql-client-core-5.1 is not installed, so not removed
Package mysql-server-5.1 is not installed, so not removed
Package mysql-server-core-5.1 is not installed, so not removed
Package mysql-source-5.1 is not installed, so not removed

MySQLi prepared statements error reporting

I'm trying to get my head around MySQli and I'm confused by the error reporting. I am using the return value of the MySQLi 'prepare' statement to detect errors when executing SQL, like this:

$stmt_test =  $mysqliDatabaseConnection->stmt_init();
if($stmt_test->prepare("INSERT INTO testtable VALUES (23,44,56)"))
{
 $stmt_test->execute();
 $stmt_test->close();
}
else echo("Statement failed: ". $stmt_test->error . "<br>");

But, is the return value of the prepare statement only detecting if there is an error in the preperation of the SQL statement and not detecting execution errors? If so should I therefore change my execute line to flag errors as well like this:

if($stmt_test->execute()) $errorflag=true;

And then just to be safe should I also do the following after the statement has executed:

if($stmt_test->errno) {$errorflag=true;}

...Or was I OK to start with and the return value on the MySQLi prepare' statement captures all errors associated with the complete execution of the query it defines?

Thanks C

Commands out of sync; you can't run this command now

I am trying to execute my PHP code, which calls two MySQL queries via mysqli, and get the error "Commands out of sync; you can't run this command now".

Here is the code I am using

<?php
$con = mysqli_connect("localhost", "user", "password", "db");
if (!$con) {
    echo "Can't connect to MySQL Server. Errorcode: %s\n". Mysqli_connect_error();
    exit;
}
$con->query("SET NAMES 'utf8'");
$brand ="o";
$countQuery = "SELECT ARTICLE_NO FROM AUCTIONS WHERE upper(ARTICLE_NAME) LIKE % ? %";
if ($numRecords = $con->prepare($countQuery)) {
    $numRecords->bind_param("s", $brand);
    $numRecords->execute();
    $data = $con->query($countQuery) or die(print_r($con->error));
    $rowcount = $data->num_rows;
    $rows = getRowsByArticleSearch("test", "Auctions", " ");
    $last = ceil($rowcount/$page_rows);
}  else {

print_r($con->error);
}
foreach ($rows as $row) {
    $pk = $row['ARTICLE_NO'];
    echo '<tr>' . "\n";
    echo '<td><a rel='nofollow' href="#" onclick="updateByPk(\'Layer2\', \'' . $pk . '\')">'.$row['USERNAME'].'</a></td>' . "\n";
    echo '<td><a rel='nofollow' href="#" onclick="updateByPk(\'Layer2\', \'' . $pk . '\')">'.$row['shortDate'].'</a></td>' . "\n";
    echo '<td><a rel='nofollow' href="#" onclick="deleterec(\'Layer2\', \'' . $pk . '\')">DELETE RECORD</a></td>' . "\n";
    echo '</tr>' . "\n";
}
function getRowsByArticleSearch($searchString, $table, $max) {
    $con = mysqli_connect("localhost", "user", "password", "db");
    $recordsQuery = "SELECT ARTICLE_NO, USERNAME, ACCESSSTARTS, ARTICLE_NAME, date_format(str_to_date(ACCESSSTARTS, '%d/%m/%Y %k:%i:%s'), '%d %m %Y' ) AS shortDate FROM AUCTIONS WHERE upper(ARTICLE_NAME) LIKE '%?%' ORDER BY str_to_date(ACCESSSTARTS, '%d/%m/%Y %k:%i:%s')" . $max;
    if ($getRecords = $con->prepare($recordsQuery)) {
        $getRecords->bind_param("s", $searchString);
        $getRecords->execute();
        $getRecords->bind_result($ARTICLE_NO, $USERNAME, $ACCESSSTARTS, $ARTICLE_NAME, $shortDate);
        while ($getRecords->fetch()) {
            $result = $con->query($recordsQuery);
            $rows = array();
            while($row = $result->fetch_assoc()) {
                $rows[] = $row;
            }
            return $rows;
        }
    }
}

I have tried reading up on this, but I am unsure of what to do. I have read about store result and free result, however these have made no difference when using them. I am unsure at exactly which point this error is being caused, and would like to know why it is being caused, and how to fix it.

Going by my debug statements, the first if loop for countQuery is not even being entered, because of an error in my sql syntax near near '% ? %'. However if I just select * instead of trying to limit based on a LIKE clause, I still get the command out of sync error.

Call to undefined method mysqli_stmt::get_result

Here's my code:

include 'conn.php';
$conn = new Connection();
$query = 'SELECT EmailVerified, Blocked FROM users WHERE Email = ? AND SLA = ? AND `Password` = ?';
$stmt = $conn->mysqli->prepare($query);
$stmt->bind_param('sss', $_POST['EmailID'], $_POST['SLA'], $_POST['Password']);
$stmt->execute();
$result = $stmt->get_result();

I get the error on last line as: Call to undefined method mysqli_stmt::get_result()

Here is the code for conn.php:

define('SERVER', 'localhost');
define('USER', 'root');
define('PASS', 'xxxx');
define('DB', 'xxxx');
class Connection{
    /**
     * @var Resource 
     */
    var $mysqli = null;

    function __construct(){
        try{
            if(!$this->mysqli){
                $this->mysqli = new MySQLi(SERVER, USER, PASS, DB);
                if(!$this->mysqli)
                    throw new Exception('Could not create connection using MySQLi', 'NO_CONNECTION');
            }
        }
        catch(Exception $ex){
            echo "ERROR: ".$e->getMessage();
        }
    }
}

If I write this line:

if(!stmt) echo 'Statement prepared'; else echo 'Statement NOT prepared';

It prints 'Statement NOT prepared'. If I run the query directly in the IDE replacing ? marks with values, it works fine. Please note that $conn object works fine in other queries in the project.

Any help please.......

Do I have to guard against SQL injection if I used a dropdown?

I understand that you should NEVER trust user input from a form, mainly due to the chance of SQL injection.

However, does this also apply to a form where the only input is from a dropdown(s) (see below)?

I'm saving the $_POST['size'] to a Session which is then used throughout the site to query the various databases (with a mysqli Select query) and any SQL injection would definitely harm (possibly drop) them.

There is no area for typed user input to query the databases, only dropdown(s).

<form action="welcome.php" method="post">
<select name="size">
  <option value="All">Select Size</option> 
  <option value="Large">Large</option>
  <option value="Medium">Medium</option>
  <option value="Small">Small</option>
</select>
<input type="submit">
</form>

PHP Commands Out of Sync error

I am using two prepared statements in PHP/MySQLi to retrieve data from a mysql database. However, when I run the statements, I get the "Commands out of sync, you can't run the command now" error.

Here is my code:

    $stmt = $mysqli->prepare("SELECT id, username, password, firstname, lastname, salt FROM members WHERE email = ? LIMIT 1";
    $stmt->bind_param('s', $loweredEmail);
    $stmt->execute();
    $stmt->store_result();
    $stmt->bind_result($user_id, $username, $db_password, $firstname, $lastname, $salt);
    $stmt->fetch();

    $stmt->free_result();
    $stmt->close();

    while($mysqli->more_results()){
        $mysqli->next_result();
    }

    $stmt1 = $mysqli->prepare("SELECT privileges FROM delegations WHERE id = ? LIMIT 1");
    //This is where the error is generated
    $stmt1->bind_param('s', $user_id);
    $stmt1->execute();
    $stmt1->store_result();
    $stmt1->bind_result($privileges);
    $stmt1->fetch();

What I've tried:

• Moving the prepared statements to two separate objects.

• Using the code:

  while($mysqli->more_results()){
      $mysqli->next_result();
  }
  //To make sure that no stray result data is left in buffer between the first
  //and second statements
  

• Using free_result() and mysqli_stmt->close()

PS: The 'Out of Sync' error comes from the second statement's '$stmt1->error'

How to start and end transaction in mysqli?

As far as I understood transaction starts once we call $mysqli->autocommit(FALSE); statement and ends after calling $mysqli->commit(); command like in the example below.

<?php
//Start transaction 
$mysqli->autocommit(FALSE);
$mysqli->query('UPDATE `table` SET `col`=2');
$mysqli->query('UPDATE `table1` SET `col1`=3;');
$mysqli->commit();
//End transaction

//Executing other queries without transaction control
$mysqli->query("Select * from table1");
$mysqli->query("Update table1 set col1=2");
//End of executing other queries without transaction control

//Start transaction 
$mysqli->autocommit(FALSE);
$mysqli->query('UPDATE `table` SET `col`=2');
$mysqli->query('UPDATE `table1` SET `col1`=3;');
$mysqli->commit();
//End transaction
?>

Have I understood correctly? If not could you please correct me, because it is actually my first time using transactions in the real life.

Thank you.