Description:
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I noticed that during the execution of an eval() the environment for an include() is kept the same as that of the original script executing the eval() but this probably often isn't the case, for example in the case of user contributed code to 'eval()'.

To overcome this problem I currently use:

$cwd=getcwd();
chdir(code-environment-of-code-to-eval);
eval(code-to-eval);
chdir($cwd);

But perhaps this could be solved in a better way?
Because in this I am assuming that the data
environment is the same as the code environment
for the code to eval().

BTW. I notice that PHP often makes no distinction
between the current working directory for code and
for data. I think that all the PHP definitions should
be rewritten to make this distinction in all relevant
cases.

BTW2. Perhaps there should even be different
workspaces for input and output data files...