关于PHP语句
有人能帮我为这句话编代码吗关于PHP语句,php,if-statement,Php,If Statement,有人能帮我为这句话编代码吗 <?php if(get_post_meta($post->ID, "price", $single = true) != ""){ ?> <?php echo get_post_meta($post->ID, "price",$single = true); ?> <?php } ?> 如果我的自定义字段为空,我希望语句继续使用此代码 <?php $asin = get_post_meta($post-&g
<?php if(get_post_meta($post->ID, "price", $single = true) != ""){ ?>
<?php echo get_post_meta($post->ID, "price",$single = true); ?>
<?php } ?>
如果我的自定义字段为空,我希望语句继续使用此代码
<?php $asin = get_post_meta($post->ID, 'asin', true);
echo azon_us_price($asin);
?>
我想您应该这样做:
<?php
if(!empty(get_post_meta($post->ID, "price", $single = true))){
echo get_post_meta($post->ID, "price",$single = true);
} else {
$asin = get_post_meta($post->ID, 'asin', true);
echo azon_us_price($asin);
}
?>
hi Null,在带有价格id的post meta中,如果价格id中为空,我希望语句继续使用第二个语句并显示结果价格id$post->id是什么?