聪明的邮政价值问题

Question

I am using the following code in my add user form:

<input type="text" size="50" name="name" 
value="{if isset($post.name)}{$post.name}{elseif isset($details.name)}{$details.name}{/if}" />

However the form returns blank in case of any error(PHP validation) but it should show post value( $post.name )

Answer 1

i have solved the issue mysql, it should be $smarty.post.name instead of $post.name , thanks for your support.