如何获取具有php代码的drupal节点的HTML输出？

Question

I have a node with the following content (input mode: php code):

this is a <?php echo "test" ?>

If I output node->body, the output is

this is a <?php echo "test" ?>

What I want is:

this is a test

What is the easiest way to do this?

(I don't want all the default divs and other structural stuff coming with it when I call node_view)

Answer 1

在Drupal中，要显示具有“ PHP筛选器”的节点的html输出，可以执行此操作

echo php_eval(node->body);

Answer 2

You should enable (shipped with core) "php.module", which comes with an input-formatter, PHP-filter. That filter allows you to insert php as you mention in nodes. And will eval() that for you on output.

Answer 3

It sounds like you don't use the php filter that's needed to do something like this. The result is your code is escaped.

On a random drupal install I just tested:

This is a <?php echo 'test'; ?>

Outputs

This is a test

When I view the node.

Answer 4

你可以输出

eval($node->body); // be sure what you are evaling