Php explode error with date and time

Question

I have converted my database from mysql to SQL server and working on exploding date and time. I am getting error: explode() expects parameter 2 to be string, This i

Answer 1

try this one:

explode(":", $r['time']->format("H:i:s"));

Answer 2

Try this,

     $time = date('H:i:s'); 
     $date = date('Y-m-d');
     echo $time."<br>";
     echo $date."<br>";
     $pieces = explode(":", $time);
     echo $pieces[0]." ".$pieces[1]." ".$pieces[2]."<br>";
     $pieces = explode("-", $date);
     echo $pieces[0]." ".$pieces[1]." ".$pieces[2];

Answer 3

You're not checking the result of your query properly.

If the second parameter of explode is marked to be no string, it's surely no string.

Thus: your sql seems not to work in the context of your PHP script, although it is proper SQL.

Check your $conn variable if it's a proper connection

dump sqlsrv error like:

$sth = sqlsrv_query($conn,"

SELECT routines.date, routines.time, 
SUM( CASE WHEN measurements.title =  'T_Luft_Temperatur' THEN CAST( REPLACE( routines.value,  ',',  '.' ) AS DECIMAL( 18, 2 ) ) ELSE NULL END) AS Temperatur
FROM routines
INNER JOIN measure_routine ON routines.id = measure_routine.routine_id
INNER JOIN measurements ON measure_routine.measure_id = measurements.id
INNER JOIN pools ON measure_routine.pool_id = pools.id
GROUP BY routines.date, routines.time
ORDER BY routines.date, routines.time;

;");

if( $sth === false ) {
     die( print_r( sqlsrv_errors(), true));
}

EDITED:

as from your last edit, the image (why an image?) shows the following error message:

explode() expects parameter 2 to be string, object given in ...

that means your variable $dateString = $r['date']; is a DateTime object. At the end you don't need to explode that, since you may access the members through the getters of this object:

$dateObj = $r['date'];
$year = $dateObj->format('Y');