How to get TOTAL DAYS from two dates (one in SQL database and then todays date (system date))

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I am not very good in php and javacript, need help on how to fix the next info,
Need to show total days in html table.

I have the first date in mysql table [CREATED_AT] and then todays date (not in database). Only what to show the total days in my html table, when I open the webpage.

Did get the script code from searching , but it does not show the info at all.

Not sure how to use print in the script

How do I get it to work

Edit: Further information that OP posted in a solution. Putting it here before it is deleted
it is giving error in

echo ("%d days\n", $days);
.
So when I take out that line then it the give me this error.

Warning: Use of undefined constant today - assumed 'today' (this will throw an Error in a future version of PHP) in C:\Mylinks\Kripto\Crypto\index.php on line 694

I do not have "today;" in my database. I just want it to get date from system date for it.

What I have tried:

<pre lang="PHP"><pre><?php $date = $checkSqlRow["DAYS_OLD"]; ?></td>
<script type="text/javascript">
$date1 = $date = $checkSqlRow["CREATED_AT"];
$date2 = today;
$diff = abs(strtotime($date2) - strtotime($date1));
$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));
echo ("%d days\n", $days);

</script>

Posted 21-Jul-22 1:25am

Odiez Bez

Updated 21-Jul-22 2:40am

CHill60

v3

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Richard MacCutchan 21-Jul-22 9:45am

See my updated solution.

1 solution

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Richard MacCutchan · Accepted Answer · 2022-07-21T01:47:00

The code you are trying to use is PHP not Javascript. Try this:
PHP

<?php $date = $checkSqlRow["DAYS_OLD"]; ?></td> <?php $date1 = $date = $checkSqlRow["CREATED_AT"]; $date2 = today; $diff = abs(strtotime($date2) - strtotime($date1)); $years = floor($diff / (365*60*60*24)); $months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24)); $days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24)); echo ("%d days\n", $days); ?>

This works (assuming $date1 is valid):
[edit]

PHP

$date2 =  date("y-m-d"); // this returns today's date
$diff = abs(strtotime($date2) - strtotime($date1));
$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));
echo $years . " years\n";
echo $months . " months\n";
echo $days . " days\n";

[/edit]