mysqli prepare statment error?

Question

Hi all,

$mysqli = new mysqli("localhost", "root", "", "test");
$mysqli->query('PREPARE mid FROM "SELECT name FROM test_user WHERE id = ?"');
// working code start
//$res = $mysqli->query('PREPARE mid FROM "SELECT name FROM test_user"  ');
//$res = $mysqli->query( 'EXECUTE mid;') or die(mysqli_error($mysqli));
// working code end..
$res = $mysqli->query( 'EXECUTE mid 1;') or die(mysqli_error($mysqli));

while($resu = $res->fetch_object()) {
    echo '<br>' .$resu->name;
}

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 '1' at line 1

my php version is PHP Version 5.3.0 and mysql

mysqlnd 5.0.5-dev - 081106 - $Revision: 1.3.2.27 $ 

I got the correct result with out using the where clause

Answer 1

Use the prepare function for a SELECT query:

http://php.net/manual/en/mysqli.prepare.php

/* create a prepared statement */
if ($stmt = $mysqli->prepare("SELECT District FROM City WHERE Name=?")) {

    /* bind parameters for markers */
    $stmt->bind_param("s", $city);

    /* execute query */
    $stmt->execute();

    /* bind result variables */
    $stmt->bind_result($district);

    /* fetch value */
    $stmt->fetch();

    /* close statement */
    $stmt->close();
}

Answer 2

http://dev.mysql.com/doc/refman/5.1/en/execute.html

You need to have a using clause in the EXECUTE statement. So you could do something like:

$mysqli = new mysqli("localhost", "root", "", "test");
$mysqli->query('PREPARE mid FROM "SELECT name FROM test_user WHERE id = ?"');
$mysqli->query('SET @var1 = 1;');
$res = $mysqli->query( 'EXECUTE mid USING @var1;') or die(mysqli_error($mysqli));