json dont working because it shows a mysql num rows warning

Question

Hello i have an annoying error in my PHP Code. I asked my teacher and he mean that the code should work and he has no idea why this error come. I hope you find the error:
In this Code i try to become a json format:

<?php
$von = $_GET['datumvon'];
$bis = $_GET['datumbis'];
$frei =0;

$DB_server = "localhost";
$DB_name = "dbterminkalenderha";
$DB_user = "root";
$DB_password = "root";

   $server = mysqli_connect($DB_server, $DB_user, $DB_password);
   $connection = mysqli_select_db($server, $DB_name);


    $myquery = $server->prepare("SELECT T_Datum,T_Frei FROM t_terminslots WHERE (T_Datum Between ? AND ?) AND (T_Frei = ?); ");
    $myquery->bind_param("ssi",$von,$bis,$frei);
    $myquery ->execute();


    //$query = mysqli_($myquery);

    if ( ! $myquery ) {
        echo mysqli_error();
        die;
    }    
    $data = array();    

    for ($x = 0; $x < mysql_num_rows($myquery); $x++) {
        $data[] = mysqli_fetch_assoc($myquery);
    }

    echo json_encode($data);     

mysqli_close($server);
?>

Error: Warning: mysql_num_rows() expects parameter 1 to be resource, object given in

Why are you mixing mysql_* functions and mysqli_* functions? ;) — Jonnix, Nov 05 '15 at 15:48
ah good to know put when i write mysqli_num_rows the same error comes — myworld, Nov 05 '15 at 15:55

score 0 · Answer 1 · answered Nov 05 '15 at 15:49

0

You need to use the mysqli function.

for ($x = 0; $x < mysqli_num_rows($myquery); $x++) {
    $data[] = mysqli_fetch_assoc($myquery);
}

answered Nov 05 '15 at 15:49

geggleto

2,605
1
15
18

json dont working because it shows a mysql num rows warning

1 Answers1