<?php
ini_set('display_errors','1');
error_reporting(E_ALL);
include_once 'dbConnect.php';
$startdate = trim($_POST['startdate']);
$enddate = trim($_POST['enddate']);
if (connect()){
global $conn;
$query="SELECT electionNo FROM election ORDER BY electionNo DESC LIMIT 1";
$details = $conn->query($query);
while ($rows = $details->fetch_assoc())
$election = $rows['electionNo'];
$election=$election+1;
$liststart= explode("T",$startdate);
$listend= explode("T",$enddate);
$start=$liststart[0]." ".$liststart[1];
$end=$listend[0]." ".$listend[1];
$year = substr($listend[0],0,4);
$insertquery = "INSERT INTO election(electionNo,year,startTime,endtime) VALUES('$election','$year','$start','$end')";
$insert = $conn->query($insertquery);
if ($insert)
echo 'Registered Successfully';
else
echo 'No good';
}
?>
我想将$start
传递给 Cron作业以安排作业。例如,如果$start = '2018-03-20 12:00:00'
,则 Cron作业应如下所示:
00 12 20 03 * php /home/Dropbox/WebServer/paramGen.php
这意味着paramGen.php
必须在2018-03-20的12:00运行。
这是否可行,如果可行,如何将$start
传递给 cron作业?
答案 0 :(得分:1)
以下是构建cron命令的方法:
$start = '2018-03-20 12:00:00';
$date=strtotime($start);
$str= date('i',$date)." ".date('H',$date)." ".date('d',$date)." ".date('m',$date)." \* php /home/Dropbox/WebServer/paramGen.php";
然后
exec("echo $str >> cronfile")
希望这会有所帮助