Send message forward along the info [closed] - php

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Send message forward along the info. the info that I select in shell or worse than for the bare lies into the database again what I need .. but the problem is such that it appears with this error:
error 1: Commands out of sync; you can't run this command now
PHP/MYSQLI
if($stmt = $this->mysqli->prepare(' SELECT id, idunik, fra, message, datoTime FROM pm WHERE id = ?' ))
{
$stmt->bind_param('i', $id);
$id = $_GET['id'];
$stmt->execute();
$stmt->bind_result($id, $idunik, $fra, $message, $datoTime);
while($stmt->fetch())
{
if ($stm = $this->mysqli->prepare('INSERT INTO pm (idunik, title, fra, til, message, datoTime) VALUES (?, ?, ?, ?, ?, NOW())')) {
$stm->bind_param('issss', $idunik, $title, $fra, $til, $message);
$idunik = $idunik;
$title = $title;
$fra = $_SESSION["id"];
$til = $fra;
$message = $_POST["tekst"];
header('Location: /besked/' . $_SESSION["id"] . '/');
$stm->execute();
$stm->close();
} else {
echo 'error 1: ' . $this->mysqli->error;
}
}
$stmt->close();
}
else
{
echo 'error 2: ' . $mysqli->error;
}
so the problem is such that when carrying it into the database then comes the error.

Answer 1

Unfortunately your question it not perfectly understandable, so I'm just going to give some general info related to the the error you receive (as I presume this is what you want to resolve):
error 1: Commands out of sync; you can't run this command now
Because Mysqli uses unbuffered queries by default on prepared statements, while Mysqli has previous results to be fetched you cannot call another procedure.
To fix this, and avoid the error you are getting, use mysqli_store_result() to get all rows first, then perform the next query.
mysqli_store_result():
http://php.net/manual/en/mysqli.store-result.php
Transfers the result set from the last query on the database
connection represented by the link parameter to be used with the
mysqli_data_seek() function.
And more info on "commands-out-of-sync":
http://dev.mysql.com/doc/refman/5.1/en/commands-out-of-sync.html