Question: ```
<?php cp_latest_tweet('<?=$next_fixture?>'); ?>

```

The `next_fixture` part is where I'm having trouble. How do I go about this?

Answer: 
Why the hell would you do that?

Simply pass the var directly:

```
<?php cp_latest_tweet($next_fixture); ?>

```

---

If you want to have a string before/after it, e.g.:

```
<?php '<div>' . cp_latest_tweet($next_fixture) . '</div>'; ?>

```

or

```
<div><?php cp_latest_tweet($next_fixture); ?></div>

```
Have you got this fixed yet, the previous two answers are correct but do you know where the variable $next\_fixture exists? if it's in the same script where you CALL the function then you should have no problem, but if it's in another script called something like next\_fixture.php you will have to include this script in the one your calling the function from. If this is the case you can use require\_once:

require\_once('next\_fixture.php'); 

and then:

cp\_latest\_tweet($next\_fixture);