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Asked: 2026-05-30T16:49:17+00:00

This is what I have for the code: <?php $link = SELECT * FROM

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This is what I have for the code:

<?php
    $link = "SELECT * FROM epk";
    echo '<b><font face="Arial, Helvetica, sans-serif" size="2" color="#FF6666">' . date("m.d.Y", strtotime($startTime)) . '</b>';
?>

I read a lot of the different fixes on this site, but I can’t seem to translate it to my line of code. I didn’t write this, but the person that did it for me has become unreachable. If anyone can tell my how to fix this I would appreciate it.

Thank you VERY much!

CORRECTION:

I’m so sorry. I have no idea why I posted that code. That was not the original one, what I posted was one I found that I tried to mess with. THIS is the original one that I have now that used to work:

<?php $link = "SELECT * FROM epk"; echo '<b><font face="Arial, Helvetica, sans-serif" size="1" color="#FF6666">' . date("m-d-y",strtotime($r['date'])) . '</b>'; ?>

Many thanks! 🙂

CORRECTION

I got it… for what I had, I just made a wild guess and added the letters “OW” to the $r line and it worked. Here it is, and hopefully it can help someone else out if they see this…

<?php $link = "SELECT * FROM epk"; echo '<b><font face="Arial, Helvetica, sans-serif" size="1" color="#FF6666">' . date("m-d-y",strtotime($row['date'])) . '</b>'; ?>

Hope that helps someone else out too.

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1 Answer

  1. Editorial Team

    It looks like your missing some information here.

    Your $startTime variable is not initialized in this code, it is never “set” anywhere. So your essentially passing “0” into your strtotime function. This means “0 seconds after the Unix Epoche, December 31, 1969 at midnight”.

    You would need to run the query called “link” first, and extract the data stored in the resulting query and pass that into the variable $startTime.

    Basically, your missing a bunch of code here.

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