CODEWITHSUNDEEP
phpmysql
dapidmini
dapidmini

Reputation: 1625

strtotime error empty time parameter when stored into a variable

I'm trying to calculate the difference in minutes between two database fields using php. the first field (attendance) is timestamp, and the second field (check_in) is time.

this is the PHP code I tried:

$sql = "
    select e.attendance, s.check_in, s.check_out
    from employee e
    inner join my_schedule s on s.id = e.sch_id
    where e.emp_id = '".$emp_id."'
";
$query = mysql_query($sql);
if ($query) {
    $row = mysql_fetch_array($query);
    echo 'employee checked in at '.strtotime($row['attendance']); // employee checked in at 2015-11-18 08:48:23  
    $late = round((strtotime($row['check_in']) - strtotime($row['attendance']))/60,0);
    if ($late < 0) {
        echo "you're late!!";
    }

the "you're late" message did appear but there's also an error message :

strtotime() [function.strtotime]: Called with an empty time parameter. in C:\Program Files\xampp\htdocs\myproject\page1.php

why do the error only appear when I store the value from strtotime() into a variable?

I'm using :

Current PHP version: 5.0.5

MySQL 4.1.14

UPDATE:

this is the result of echo var_dump($row['attendance']); (data type timestamp)

string(19) "2015-11-18 08:48:23"

and this is echo var_dump($row['check_in']); (data type time)

string(8) "08:00:00"

is it normal that PHP perceives those fields as string instead of time?

UPDATE 2 : I found the source of the problem! the query used a parameter (emp_id) and one of the emp_id returns a zero result so the strtotime returned the error.

the solution is that I added an IF clause to make sure that the rows have a valid value before running strtotime.

if ($row['check_in'] != '' && $row['attendance'] != '') {
    $late = round((strtotime($row['check_in']) - strtotime($row['attendance']))/60,0);
}

Upvotes: 1

Views: 1049

Answers (2)

jaggedsoft
jaggedsoft

Reputation: 4038

The issue here is that strtotime expects a date, not just a time. So you can extract the date from your attendance like this:

$attendance = "2015-11-18 08:48:23";
$check_in = "08:00:00";
if ( strlen($check_in) <= 8 ) { // Does not include date. Prepend it automatically:
    $parts = explode(" ",$attendance);
    $attendance_date = $parts[0];
    $check_in = "{$attendance_date} {$check_in}";
}
$late = round((strtotime($check_in) - strtotime($attendance))/60,0);

Just to be clear before we run our calculations we want to end up with:

$attendance = "2015-11-18 08:48:23";
$check_in = "2015-11-18 08:00:00";

Here are some more examples of strtotime()

echo strtotime("2015-11-18 08:48:23"),"\n";
echo strtotime("18 November 2015"),"\n";
echo strtotime("now"),"\n";
echo strtotime("+1 day"),"\n";
echo strtotime("+1 week"),"\n";
echo strtotime("+1 week 2 days 3 hours 4 seconds"),"\n";
echo strtotime("last Monday"),"\n";
echo strtotime("next Friday"),"\n";

Read more: php.net/strtotime

Upvotes: 2

Ajay Makwana
Ajay Makwana

Reputation: 2372

if this is comes in form of warning then put @ before the strtotime().

$late = @strtotime($row['check_in']) - @strtotime($row['attendance']);

its just hide warnings.

Upvotes: -1

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