Question

我正在尝试使用此代码从数据库创建日期创建24小时的时间段，但是当我想要回显时，我有mysql_fetch_assoc问题，它显示错误的格式我只需要几分钟！

//connect to database
//.... 

//get time difference in seconds from last execution
$sql1 = "SELECT TIME_TO_SEC(TIMEDIFF(NOW(), last_ts)) AS tdif FROM php_cron";
$res1 = mysql_query($sql1) or die("[1] MySQL ERROR: ".mysql_error());


while ($dif = mysql_fetch_assoc($res1)) {

echo $dif["tdif"] ;

}

if ($dif >= 86400) { //24h

    //following code will run once every 24h

    //update user's page rank
    $sql2 = "UPDATE myTable SET `user-rank` = `user-rank` + 10"; 
    mysql_query($sql2) or die("[2] MySQL ERROR: ".mysql_error());

    //update last execution time
    $sql3 = "UPDATE php_cron SET last_ts = NOW() WHERE id=1";
    mysql_query($sql3) or die("[3] MySQL ERROR: ".mysql_error());

}

如果您可以访问列的所有日期会更好:)任何人都可以帮忙吗？

=============================================== ========================================= ///////// ///

我把ID放在柜台上。还有其他更好的方法吗？感谢

$idcount = 1;
while ($dif = mysql_fetch_assoc($res1)) {
echo $dif["tdif"];
echo "<br>";

if ($dif["tdif"] >= 86400) { //24h
  //update user's page rank
  $sql2 = "UPDATE deposit SET `earn` = `earn` + 10 where id=" . $idcount . ""; 
  mysql_query($sql2) or die("[2] MySQL ERROR: ".mysql_error());

Answer 1

我的猜测这实际上就是你想要做的。请注意，我已将if语句移至while循环内，并将$dif更改为$dif["tdif"]。

$sql1 = "SELECT TIME_TO_SEC(TIMEDIFF(NOW(), last_ts)) AS tdif FROM php_cron";
$res1 = mysql_query($sql1) or die("[1] MySQL ERROR: ".mysql_error());

while ($dif = mysql_fetch_assoc($res1)) {
   echo $dif["tdif"];

   if ($dif["tdif"] >= 86400) { //24h
      //update user's page rank
      $sql2 = "UPDATE myTable SET `user-rank` = `user-rank` + 10"; 
      mysql_query($sql2) or die("[2] MySQL ERROR: ".mysql_error());

      //update last execution time
      $sql3 = "UPDATE php_cron SET last_ts = NOW() WHERE id=1";
      mysql_query($sql3) or die("[3] MySQL ERROR: ".mysql_error());

   }
}

我正在创造24小时期间价值。不要工作mysql_fetch_assoc

1 个答案: