假设你数据库中的时间字段为news_time
<?php do { ?>
<TR>
<TD>
<?php echo $row_gzap['news_title']; ?>
<?php if( (time()-strtotime($row_gzap['news_time']))/(1000*3600*24)<30 ) {?>
<img src="images/new_pa20.gif" >
<?php } ?>
</TD>
</TR>
<?php } while ($row_gzap = mysql_fetch_assoc($gzap)); ?>
<?php do { ?>
<TR>
<TD>
<?php echo $row_gzap['news_title']; ?>
<?php if( (time()-strtotime($row_gzap['news_time']))/(1000*3600*24)<30 ) {?>
<img src="images/new_pa20.gif" >
<?php } ?>
</TD>
</TR>
<?php } while ($row_gzap = mysql_fetch_assoc($gzap)); ?>
echo"$news_tile";
echo"图片";
TO: wen8u8
你这样我试了也不行,这行会报错,
另 echo"图片"; 这样不行吧,我图片没会在库里作为一个字段的,
总不能<?php echo <img src="images/new_pa20.gif"> ?>吧~~。
哪位高手指点下下
$t="2005-08-01";
echo pic($t);
function pic($date)
{
preg_match('/(\d{4})\-(\d{2})\-(\d{2})/',$date,$setdate);
if ((time() - mktime(0,0,0,$setdate[2],$setdate[3],$setdate[1])) < 2*24*60*60) $pic = "<img src='./new.gif'>";
return $pic;
}
?>
$img = '<img src="images/new_pa20.gif" >';
if(strtotime("$row_gzap[news_time] -2 month") < time()) $img = '';
?>
<TR>
<TD>
<?php echo $row_gzap['news_title'].$img; ?>
</TD>
</TR>
<?php } while ($row_gzap = mysql_fetch_assoc($gzap)); ?>ice_berg16(寻梦的稻草人) 代码的问题在于把unix时间戳的计量单位弄错了,php是“秒”而js才是毫秒
if(strtotime("$row_gzap[news_time] + 2 month") > time()) $img = '';
<?php do { ?>
<TR>
<TD ><?php echo $row_jyhc['news_title']; ?>
<?php if(strtotime("$row_jyhc[news_time] +2 month") > time() ) { ?>
(<?php echo $row_jyhc['news_time']; ?> )
<img src="../images/new.gif" >
<?php } ?>
</TD>
</TR>
<?php } while ($row_jyhc = mysql_fetch_assoc($jyhc)); ?>