php - mysql_fetch_assoc() expects parameter 1 to be resource, null given in -
i follow topic because have same problem ( can't use command shell, edit file host ) -> change value after 24 hours
first run sql
create table `php_cron` ( `id` int(11) unsigned not null auto_increment, `last_ts` datetime default null, primary key (`id`) ); insert `php_cron` (`id`, `last_ts`) values (1,'2012-08-10 00:00:00');
and code
$res1 = mysql_query("select time_to_sec(timediff(now(), last_ts)) tdif php_cron id=1"); $dif = mysql_fetch_assoc($dif['tdif']); if ($dif >= 86400) { //24h //following code run once every 24h //update user's page rank $sql2 = "update logs_limitbandwidthtoday set bandwidthtoday = 0"; mysql_query($sql2); $sql23 = "update logs_limitlinktoday set limitlink = 0"; mysql_query($sql23); $sql24 = "update logs_limitvipbw set bandwidthtoday = 0"; mysql_query($sql24); $sql25 = "update logs_limitviplink set limitlink = 0"; mysql_query($sql25); $sql26 = "update account_vip set alltime = alltime - 1 alltime > 0"; mysql_query($sql26); //update last execution time $sql3 = "update php_cron set last_ts = now() id=1"; mysql_query($sql3); }
error -> php warning: mysql_fetch_assoc() expects parameter 1 resource, null given in /.... on line 2
and not sure code still work or not, please give me answer question. thank !
ok should solve problem
$dif = mysql_fetch_assoc($res1); $dif1 = $dif['tdif']; if ($dif1 >= 86400) {
but since mysql_* outdated , removed php 7.0 updated code should this
$res1 = mysqli_query($con,"select time_to_sec(timediff(now(), last_ts)) tdif php_cron id=1"); $dif = mysqli_fetch_assoc($res1); $dif1 = $dif['tdif']; if ($dif1 >= 86400) { //following code run once every 24h //update user's page rank $sql2 = "update logs_limitbandwidthtoday set bandwidthtoday = 0"; mysqli_query($con,$sql2); $sql23 = "update logs_limitlinktoday set limitlink = 0"; mysqli_query($sql23); $sql24 = "update logs_limitvipbw set bandwidthtoday = 0"; mysqli_query($con,$sql24); $sql25 = "update logs_limitviplink set limitlink = 0"; mysqli_query($con,$sql25); $sql26 = "update account_vip set alltime = alltime - 1 alltime > 0"; mysqli_query($con,$sql26); //update last execution time $sql3 = "update php_cron set last_ts = now() id=1"; mysqli_query($con,$sql3); }
where $con connection sting make database
$server='localhost';//your host name $user_name='root';//your mysql user name root default $password='';//your mysql password blank default $database='db';//your database name $con=mysqli_connect($server,$user_name,$password,$database);
Comments
Post a Comment