数据从一个Php页面传递到另一个页面

<a href="<?php echo $config_basedir;?>showcart.php?type=" . $type . >View   Basket/Checkout</a>

Above mentioned link is called by header program . I want to pass the type value while calling the showcart.php. But its passing blank value.

EDIT

<?php $type=$_GET['type']; 
echo "$type"; ?> 
<a href="<?php echo $config_basedir; ?>">Home</a>
 - <a href="<?php echo $config_basedir;?>showcart.php?type=" . $type . >View Basket/Checkout</a> 
</div>

When you try to throw in the $type variable you do so from within the HTML context. You need to open php tags again and echo the $type variable where you need to

In result you will get:

 <a href="<?php echo $config_basedir;?>showcart.php?type=<?php echo $type; ?>">View   Basket/Checkout</a>

You might as well just do it all as one echo:

echo "<a href='{$config_basedir}showcart.php?type={$type}'>View  Basket/Checkout</a>";

As you have it there are several syntax errors, such as no closing quote on your href attribute, and you are trying to concatenate with PHP outside of a PHP block.