date output format

[sirTemplar]

EVENTDATE is a date field (yyyy-mm-dd). if i echo/print it on the output with this code:

{$row['EVENTDATE']}

it will give me something like: 1972-07-27

what if i want to show on the output only the year i.e. 1972? what should i try to put?

thanks

[laserlight]

The simplest solution would be to

echo substr($row['EVENTDATE'], 0, 4);

[sirTemplar]

i tried what you suggest and is giving me this error:

Parse error: parse error, expecting T_STRING' orT_VARIABLE' or `T_NUM_STRING' in /var/add/addl/websites/site/site/list.php on line 67

let me give you all the present echo:

echo "<table bgcolor=#FFFEEF border=0 cellpadding=2 cellspacing=0 style=border-collapse: collapse bordercolor=#111111 width=100%>
<tr><td align=left width=20% valign=top><b><font face=Verdana size=2 color=#800080><li>{$row['EVENTDATE']}</font></b></td>
<td align=justify width=80% valign=top><font face=Verdana size=2 color=#000000>{$row['EVENTTEXT']}</td></table>";

the ebove works ok only that the EVENTDATE is complete, when i put the substring, it gives me that error!

🙁

[laserlight]

echo "<table bgcolor=#FFFEEF border=0 cellpadding=2 cellspacing=0 style=border-collapse: collapse bordercolor=#111111 width=100%>
<tr><td align=left width=20% valign=top><b><font face=Verdana size=2 color=#800080><li>" . substr($row['EVENTDATE'], 0, 4) . "</font></b></td>
<td align=justify width=80% valign=top><font face=Verdana size=2 color=#000000>{$row['EVENTTEXT']}</td></table>";

[sirTemplar]

thank you, that did the job! 😃