Question

PHP判断语句

如果$goods.seller_note值不为空，执行
<a href="{$goods.seller_note}" target="_blank"><img src="images/bnt_cat.gif" border="0"
/></a>

如果$goods.seller_note值为空，则执行

<a href="javascript:addToCart({$goods.goods_id})"><img src="images/bnt_cat.gif" /></a>

这个判断语句应该怎样写啊？

Answer 1

<!--{if $goods.seller_note eq ''}-->
<a href="javascript:addToCart({$goods.goods_id})"><img src="images/bnt_cat.gif" /></a>
<!--{else}-->
<a href="{$goods.seller_note}" target="_blank"><img src="images/bnt_cat.gif" border="0"
/></a>
<!--{/if}-->

Answer 2

<?php
if($goods.seller_note) {
?>
<a href="<?php echo $goods.seller_note; ?>" target="_blank"><img src="images/bnt_cat.gif" border="0"
/></a>
<?php
} else {
?>
<a href="javascript:addToCart(<?php echo $goods.goods_id; ?>)"><img src="images/bnt_cat.gif" /></a>
<?php
}
?>

Answer 3

if(!empty($goods.seller_note))($goods.seller_note!="")
{?>
<a href="{$goods.seller_note}" target="_blank"><img src="images/bnt_cat.gif" border="0"
/></a>
<?php }
esle
{?>
<a href="javascript:addToCart({$goods.goods_id})"><img src="images/bnt_cat.gif" /></a>
<?php }<?>