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我按照這個話題給出空,因爲我有同樣的問題(不能使用命令行shell,只需編輯文件主機) - >change a value after 24 hoursmysql_fetch_assoc()預計參數1是資源,在
首先運行SQL
CREATE TABLE `php_cron` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`last_ts` datetime DEFAULT NULL,
PRIMARY KEY (`id`)
);
INSERT INTO `php_cron` (`id`, `last_ts`) VALUES (1,'2012-08-10 00:00:00');
而且我的代碼
$res1 = mysql_query("SELECT TIME_TO_SEC(TIMEDIFF(NOW(), last_ts)) AS tdif FROM php_cron WHERE id=1");
$dif = mysql_fetch_assoc($dif['tdif']);
if ($dif >= 86400) { //24h
//following code will 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);
$sql26 = "UPDATE account_vip SET ALLTime = ALLTime - 1 WHERE ALLTime > 0";
mysql_query($sql26);
//update last execution time
$sql3 = "UPDATE php_cron SET last_ts = NOW() WHERE id=1";
mysql_query($sql3);
}
錯誤 - > PHP的警告:mysql_fetch_assoc()預計參數1是資源,在空給出/ ....第2行
我不確定此代碼是否仍然有效,請給我這個問題的答案。非常感謝 !
$ dif = mysql_fetch_assoc($ res1); –
沒有$ res1 ['tdif'] - > ['tdif'] < - bro? –
這可能會工作$ dif = mysql_fetch_assoc($ res1); $ dif1 = $ dif ['tdif']; if($ dif1> = 86400){ –