How to write "if something -> print" "else if nothing -> print nothing"

Question

I am writing a php function for wordpress that is executed through an XML feed. Therefore we are excepting a feed and then based on the nodes placing those in our website. What I need help with is we have a bunch of different images of credentials (i.e BBB, chamber of commerce etc) What I need therefore is when there is a link to a BBB then it should display a picture, if not then it should be blank. The problem I am running into is because the BBB links will be random based on different businesses. Any help would be greatly appreciated. Thanks.

If URL "pic" else "no pic"

Answer 1

Do you mean this? Otherwise please explain your problem better.

if (!empty($url)) {
    echo '<img src="' .$url. '" />';
}
else {
    echo ' ';
}

Check here when empty returns false (and therefore !empty is true) and really consider if this fits your needs.

Answer 2

Maybe I'm missing something, but wouldn't this do?

if($license1) { print "<img src=\"/path/to/bbb.logo\" alt=\"BBB Logo\" />"; }

Answer 3

A method would be creating an array like $feeds = array("pic","xml"); then testing if its in array like

if in_array($url,$feeds)
// your code;

or the second method would be creating an temp var like $tmp = $url =="pic" ? "pic" : "nopic"; or to set just an boolean $tmp = $url =="pic" ? TRUE : FALSE;

then you can test it like this

if($url) // if its == "pic" it would return true otherwise false
 //make your url 

Answer 4

Thank you for all your help - I am sort of new with php + XML, but the "!empty" function is what I was looking for. Thanks

Answer 5

Also a shorthand way is to do

But when the given var is an array I think you need to use is_array(), or if it's a class objec,t use is_object() to verify that it has content.