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coding problem on php auto send mail after a period

hi all, im trying to develop a webased helpdesk, it will auto send a email to engineer after 1 day, 3 day, 5 day & 7day if the status of the report is open. both my crontab and php mail function are working well. now my problem is no idea how to set the condition in php about the date it will send email after the period. hope someone can help me and very appreciate with that, thanx alot.

Hey, not too sure on your problem, I don't quite understand what these reports are :S, but you may need to have a read up on the php date function @ http://php.net/date.
Below is a simple program I have wrote for you to demonstrate how to work out the previous date.

PHP Code:




<?php 
$date=date(Ymd); 
echo $date . '<br />'; 
echo $date - 1 . '<br />'; 
echo $date -2 . '<br />'; 
echo $date + 1 . '<br />'; 
echo $date + 2 . '<br />'; 
echo 
?>

This works best if you specify the date in the format above, Year month day. This is because you can easily subtract a day of or add a day if it's at the end of the string.

There are some problems with this however, if you look at the output:

Code:

You can see here that it is saying that the date + 2 will be the 32nd of august, which is an imposibility. Is there a php function that does this more succesffully anyone?

hi tomj88, thanx u so much, u did help me a lot :)
im facing the problem too which the date will not move into nest month and it become 32th after adding the date. hey guyz do u know what to do on this problem, plz give me a hand on it :)

I suppose you could make some huge switch case or if else statement that checks this, basically make a calender, but this may become highly complicated...

I haven't coded in php much recently, so the syntax is probably wrong...

If you build up an array of the months and how many days they have, you can address the arrays later to shorten the code.

PHP Code:




$31daymonths = ["January", "March", "May", "July", "August", "October", "December"]; 
$30daymonths = ["April", "June", "September", "November"]; 
 
if($year % 4 == 0){ 
   $29daymonth = "April"; 
}else{ 
   $28daymonth = "April"; 
}

I'm not sure about this last bit, I always forget how to use the remainder syntax. Basically, If the current year / 4 does not give us a fraction, then april has 29 days.

I would say it would be best to run a for or do while or while loop when you are adding, and only add 1 each time (or reduce). This would be better because instead of going right over, say up to day 35, you could check on each itteration if the day is too high, and then adjust accordingly.

I hope this helps... I will try to build a simple callender later on tonight if you don't manage it now ;). Keep me updated on how its going.

Quote:

[i]from http://uk.php.net/date
z The day of the year (starting from 0) 0 through 365

If you used this instead, then it would be far easier, just add a check in to see if its from the same year, and whether the year is a leap year.

thanx so much. im trying now, hope everything go well :)

This doesn't solve your problem, but it shows you how you could possibly solve it.

PHP Code:




<?php 
        # value to add 
    $value = 150; 
        # day (0 to 365) 
    $day = date('z'); 
        # the year 
    $year = date('Y'); 
 
        # check to see if it is a leap year 
    if ($year % 4 == 0){ 
        $leap = 1; 
    }else{ 
        $leap = 0; 
    } 
 
        # works out whether after adding the value 
        # it is a new year 
    if ($leap == 1){ 
        if (($day + $value) > 365){ 
            $newday= ($day + $value); 
            $newday =- 365; 
            $newyear = ++$year; 
        }else{ 
            $newday= ($day + $value); 
            $newyear = $year; 
        } 
    }else{ 
        if (($day + $value) > 364){ 
            $newday = ($day + $value); 
            $newday =- 365; 
            $newyear = ($year + 1); 
        }else{ 
            $newday = ($day + $value); 
            $newyear = $year; 
        } 
    } 
 
       # echo the data 
echo $day . " & " . $newday . '<br />'; 
echo $year . " & " . $newyear . '<br />'; 
 
?>

Hope this helps!

thanks for ur help. i just found a code, i think this can help me out, hopefully :D

PHP Code:




<? 
$tommorrow = time() + (1 * 24 * 60 * 60); 
$nextWeek = time() + (7 * 24 * 60 * 60); 
$threedays = time() + (3 * 24 * 60 * 60); 
$fivedays = time() + (5 * 25 * 60 * 60); 
                   // 7 days; 24 hours; 60 mins; 60secs 
echo 'Now:          '. date('Y-m-d') ."<br>"; 
echo 'Next Week:    '. date('Y-m-d', $nextWeek) ."<br>"; 
echo 'Tommorrow:    '. date('Y-m-d', $tommorrow) ."<br>"; 
echo 'After 3 days: '. date('Y-m-d', $threedays) ."<br>"; 
echo 'After 5 days: '. date('Y-m-d', $fivedays) ."<br>"; 
 
?>

this is the output

Quote:

Now: 2005-02-28
Next Week: 2005-03-07
Tommorrow: 2005-03-01
After 3 days: 2005-03-03
After 5 days: 2005-03-05

anyway tomj88, really thanks on trying to help me :)

edit: Removed this post, its the same as next but I had a little problem in the code in this one, and the second one I though I clicked edit but I must have clicked new post... sorry!

I was bored so I tried to implement this for you. I think you should be able to understand this, and I hope that it fulfils your needs. I'm sure there are better ways of doing this, just a quick little hack :)

PHP Code:




<?php 
 
# a function to check a report: 
# will send an email to an admin if a report has 
# been open for a day, 3 days, 5 days or anything 
# greater than 7 days 
 
# syntax: checkreport("A simple report", 365, 2005); 
# first is the name of the report, next is the day the report 
# was opened on, and the next is the year it was opened on 
# note, all days here are in z format i.e. 0 to 365 
function checkreport($reportname, $openday, $openyear){     
     
    # get the date in the format 0 to 365 
    $day = ('z'); 
    # get the year 
    $year = ('Y'); 
       
    # is it a leap year? 
    if ($year % 4 == 0){ 
        $leap = 1; 
    }else{ 
        $leap = 0; 
    } 
     
    # after how many days should the emails be sent? 
    $emaildays = array(1, 3, 5, 7); 
    $emaildayscount = count($emaildays); 
 
    # Checks once you add the $emailday to the $openday that 
    # the number is still within the 0 to 365 limit of a year. 
    # checks for leap years, and uses a for loop to itterate over the 
    # amount of days listed in $emaildays (uses $emaildayscount) 
    for($i = 0; $i < $emaildayscount; $i++){ 
        $value = $emaildays[i]; 
        if ($leap == 1){ 
            if (($openday + $value) > 365){ 
                $newopenday[i] = $openday + $value; 
                $newopenday[i] =- 365; 
                $newyear[i] = $year +1; 
            }else{ 
                $newopenday[i] = $openday + $value; 
                $newyear[i] = $year; 
            } 
        }else{ 
            if (($openday + $value) > 364){ 
                $newopenday[i] = $openday + $value; 
                $newopenday[i] =- 364; 
                $newyear[i] = $year + 1; 
            }else{ 
                $newopenday[i] = $openday + $value; 
                $newyear[i] = $year; 
            } 
        } 
    } 
 
    # run a for loop to check if an email has to be sent 
    for($i = 0; $i < $emaildayscount; $i++){ 
         
    if($day+$emaildays[i]>=$newopenday[i]){ 
            email($reportname); 
            break; 
        # if say, it is the start of a new year,  
        # then the above might not send the email 
        # this checks if $openyear was before but 
         # not equal to this $year 
        }else if($openyear<$year){ 
            email($reportname); 
            break; 
        } 
    } 
} 
 
# replace this function with a command that sends an email 
function email($reportname){ 
    echo $reportname . " needs checking!"; 
} 
# run a check 
checkreport("A simple report", 0, 2005); 
?>

been trying whole day, finally get something :D now i using crontab run this page 10am everyday.

PHP Code:




<? 
 
// receiver for case still not close after 1 day  
$to1  = 'dyer607@hotmail.com';  
 
// receiver for case still not close after 3 days  
$to3  = 'dyer607@hotmail.com';  
 
// receiver for case still not close after 5 days  
$to5  = 'dyer607@hotmail.com';  
 
// receiver for case still not close after one week  
$to7  = 'dyer607@hotmail.com';  
 
// subject for email 
$subject = 'Problem still not solve'; 
 
 
// To send HTML mail, the Content-type header must be set 
$headers = 'From: [email]dyer607@hotmail.com[/email]' . "\r\n" . 
   'Reply-To: [email]dyer607@hotmail.com[/email]' . "\r\n" . 
   'X-Mailer: PHP/' . phpversion(); 
 
//define the period 
$oneday = time() - (1 * 24 * 60 * 60); 
             //1 day ; 24 hours ; 60 mins ; 60 secs 
$nextweek = time() - (7 * 24 * 60 * 60); 
                     //7 days ; 24 hours ; 60 mins ; 60 secs 
$threedays = time() - (3 * 24 * 60 * 60); 
             //3 days ; 24 hours ; 60 mins ; 60 secs 
$fivedays = time() - (5 * 25 * 60 * 60); 
                     //5 days ; 24 hours ; 60 mins ; 60 secs 
 
 
?> 
 
<? 
 
$connection = mysql_connect("localhost", "root", "root"); 
 
 
 
$db = "reminder"; 
 
 
 
mysql_select_db($db, $connection) or die( "Could not open $db"); 
 
$sql = "SELECT * FROM reminderdate"; 
 
$result = mysql_query($sql, $connection) or  
 
            die("Could not execute sql: $sql"); 
 
$num_result = mysql_num_rows($result); 
 
for ($i=0; $i < $num_result; $i++) 
 
{ 
 
  $row = mysql_fetch_array($result); 
 
  $id = $row["reminderid"]; 
  if ($row["casedate"] == date('Y-m-d', $oneday)) 
  { 
        // message for case after 1 day 
    $message1  = '<html><body><p>Service Report after 1 day</p><table cellspacing=3 cellpadding=3>'; 
       $message1 .= '<tr><th>ReportID</th><td>' .$row["reminderid"]; 
    $message1 .= '</td></tr><tr><th>Case Date</th><td>' .$row["casedate"];  
    $message1 .= '</td></tr><tr><th>Location</th><td>Netrixs</td>'; 
    $message1 .= '</tr><tr><th>Assigner</th><td>Joe</td></table></body></html>'; 
    // Mail it 
    mail($to1, $subject, $message1, $headers);   
  } 
  elseif ($row["casedate"] == date('Y-m-d', $threedays)) 
  { 
        // message for case after 3 days 
    $message3  = '<html><body><p>Service Report after 3 days</p><table cellspacing=3 cellpadding=3>'; 
       $message3 .= '<tr><th>ReportID</th><td>' .$row["reminderid"]; 
    $message3 .= '</td></tr><tr><th>Case Date</th><td>' .$row["casedate"];  
    $message3 .= '</td></tr><tr><th>Location</th><td>Netrixs</td>'; 
    $message3 .= '</tr><tr><th>Assigner</th><td>Joe</td></table></body></html>'; 
    // Mail it 
    mail($to3, $subject, $message3, $headers);   
  } 
  elseif ($row["casedate"] == date('Y-m-d', $fivedays)) 
  { 
        // message for case after 5 days 
    $message5  = '<html><body><p>Service Report after 5 days</p><table cellspacing=3 cellpadding=3>'; 
       $message5 .= '<tr><th>ReportID</th><td>' .$row["reminderid"]; 
    $message5 .= '</td></tr><tr><th>Case Date</th><td>' .$row["casedate"];  
    $message5 .= '</td></tr><tr><th>Location</th><td>Netrixs</td>'; 
    $message5 .= '</tr><tr><th>Assigner</th><td>Joe</td></table></body></html>'; 
    // Mail it 
    mail($to5, $subject, $message5, $headers);   
  } 
  elseif ($row["casedate"] == date('Y-m-d', $nextweek)) 
  { 
        // message for case after one week 
    $message7  = '<html><body><p>Service Report after 7 days</p><table cellspacing=3 cellpadding=3>'; 
       $message7 .= '<tr><th>ReportID</th><td>' .$row["reminderid"]; 
    $message7 .= '</td></tr><tr><th>Case Date</th><td>' .$row["casedate"];  
    $message7 .= '</td></tr><tr><th>Location</th><td>Netrixs</td>'; 
    $message7 .= '</tr><tr><th>Assigner</th><td>Joe</td></table></body></html>'; 
     
    // Mail it 
    mail($to7, $subject, $message7, $headers);   
  } 
   
} 
?>

this is the sample output i get from the email

Quote:

Service Report after 3 days
ReportID 2
Case Date 2005-08-29
Location Netrixs
Assigner Joe
Date 2005-09-01

cool, glad to here you got it working mate!

Tom