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i have function below warning $tab variable says not defined. how can define , not receive warning anymore?
<?php /*** suggested articles random ***/ function doarticle_suggested_small_horizontall($articleid,$title,$photo,$parentid,$catid,$altdescription) { $tab .= "<table width=150 cellspacing=5 style=border: 1px solid #0066ff align=right>\n"; $tab .= "<tr>\n"; $tab .= "<td align=center bgcolor=#ffffff><a href='../artandculture/adetails.php?articleid=$articleid&parentid=$parentid&catid=$catid'> <img src='../images/simage/$photo' border='0' alt='$altdescription'></a></td>\n"; $tab .= "</tr>\n"; $tab .= "<tr align=right width=150 height=80 border=0 style=border: 1px solid #ffffff>\n"; $tab .= "<td width=110 align=right dir='rtl' border=0 style=border: 1px solid #ffffff ><p class=articletitlenounderline><a href='../artandculture/adetails.php?articleid=$articleid&parentid=$parentid&catid=$catid'><strong>$title </strong></p></a></td>\n"; $tab .= "</tr>\n"; $tab .= "</table> <p> <hr class='hr99' ></hr></p>"; return $tab; } $tab = ""; ?>
you need define $tab @ beginning of function.
replace $tab .= "<table width=15 ....
with $tab = "<table width=15....
or add $tab = ""; first line of function, defining outside function should remove.
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