如果一个值存储在数据库中，我可以回显它，但如果没有存储值，我怎么能省略整个标记？

Question

echo "<a href=\"{$row['link']}\" class=\"button\">Website link</a>";

This will display an an anchor tag with a class of "button" and the href value will be whatever is in the database. If I don't have a link stored in the database, I want nothing to appear in the rendered html, but with this example the button will still show up, it just won't have an href value.

I've tried doing this:

echo "{$row['link']}";

And in the database, the 'link' value would be:

<a href="#" class="button">Website link</a>

My issue with this is that I don't like storing and pulling html code from the database. Also, if I leave the link value empty in the database, part of the button is still visible because of the "button" class.

Answer 1

It sounds like all you need is a very simple conditional.

You have a column in your table for the link so regardless of whether it has a value or not, it will be in the array.

Add an if statement to check it.

if ( $row['link'] ) {
    echo "<a href=\"{$row['link']}\" class=\"button\">Website link</a>";
}