Output of php script needs to be save in $content - php

Question

How do I need to adjust the below php if the output of the php-script needs to be saved in $content?
<?php
//db verbindung
mysql_connect("Hostname", "Username", "Password");
mysql_select_db("Database Name");
//db abfrage
$query = "SELECT
YEAR(datetime) AS dy,
MONTH(datetime) -1 AS dm,
DAY(datetime) AS dd,
HOUR(datetime) AS th,
MINUTE(datetime) AS tm,
temp,
hum,
pressure
FROM wettertabelle
WHERE DATE(datetime) = '2013-11-25'
ORDER BY datetime
";
// NEU: Variable definieren
$zeilenzaehler = 1;
//ausgabe der zeilen
$result = mysql_query($query)
OR die("Error: $query <br>" . mysql_error());
while ($row = mysql_fetch_array($result)) {
// echo
if ($zeilenzaehler != 1) {
echo ",";
}
echo "{date: new Date(" . $row['dy'] . "," . $row['dm'] . "," . $row['dd'] . "," . $row['th'] . "," . $row ['tm'] . "),t:" . $row['temp'] . ",h:" . $row['hum'] . ",p:" . $row['pressure'] . "}";
//Variable jetzt auf 2
$zeilenzaehler = 2;
};
?>
What is working is e.g.:
<?php
$content = This is the output!!
?>
or
<?php
ob_start();
echo 'This is the output!!';
$content=ob_get_contents();
ob_end_clean();
?>
How would I need to adjust the above script to get it into the following output structure:
<?php
class user_klasse
{
function ausgabe()
{
return 'This test is generated by php.';
}
}
?>

Answer 1

You only have two echo statements, change them to save the value in your variable. No need for output buffering for something so small.
if ($zeilenzaehler != 1)
{
$content.= ","; // instead of echo, and same for the one below
}
$content.= "{date: new Date(".$row['dy'].",".$row['dm'].",".$row['dd'].",".$row['th'].",".$row ['tm']."),t:".$row['temp'].",h:".$row['hum'].",p:".$row['pressure']."}";
At the end you can just do
echo $content;