loading PHP class body from external file

Question

I want to include an external file as a class body, something like this:

//$key is alphanumeric
$className='_Module_'.$key;
if(!class_exists($className))
{
    eval(' class '.$className.' extends ModuleContext');
    {
        ...include the file here...
    }
}
eval('$instance=new '.$className.'();');

This does what I want:

//$key is alphanumeric
$className='_Module_'.$key;
if(!class_exists($className))
{
    //There's got to be a better way to do this
    $contents=rtrim(ltrim(trim(file_get_contents($fileName.'.php',true)),'<?php'),'?>');
    //All on one line to preserve line numbering for errors
    eval(' class '.$className.' extends ModuleContext{ '.$contents.' }');
}
eval('$instance=new '.$className.'();');

I don't like using eval to include the whole file because it generates confusing error messages:

PHP Fatal error:  Call to undefined function doFoo() in test/test.php(57) : eval()'d code on line 7

and I've read in other posts that eval()'d code doesn't take advantage of php accelerators. Also, it requres some trimming to remove the '<?php' and '?>' tags, since they don't work so well inside a class definition.

The included file needs to be 'user friendly' to someone who is familiar with PHP, but not object oriented programming - it will have a few pre-defined functions to be filled in by the module implementer.

Is there a way to do this that doesn't use eval for the include?

@user492983 - First, you shouldn't veal code (it's risky and will be hard to read later). Second, why would you want to dynamically create classes, that's not how OOP was meant to work. Third, there's another question where the person wanted to extends a "variable" class: http://stackoverflow.com/questions/304509/variable-variable-class-extensions-in-php-is-it-possible — Francois Deschenes
– Francois Deschenes, Commented Jun 24, 2011 at 4:56
But the question is why would you want to dynamically create classes? I'm not sure that this is what OOP is all about. — Francois Deschenes
– Francois Deschenes, Commented Jun 24, 2011 at 19:59
I agree on the eval point. That's why I was looking for another solution. Apparently there is no other way to load the class body from a second file, so I've had the project requirements changed to allow the whole class in one file (which is my preference anyway). The classes aren't dynamically created, they are dynamically loaded. In this case, the classes to be loaded aren't known until run time. — Jon Hulka
– Jon Hulka, Commented Jun 24, 2011 at 20:13
@Jon - In that case $unknown = new {$class}; but that doesn't allow a class to dynamically extend another one. Dynamic means at runtime which means that you're ultimately creating a new class if you extend at runtime. — Francois Deschenes
– Francois Deschenes, Commented Jun 24, 2011 at 22:36
I see your point. That is one reason to get away from eval(). If it could have been done with an include, then conceptually it wouldn't have been dynamically extended. In reality, everything in PHP happens dynamically. — Jon Hulka
– Jon Hulka, Commented Jun 25, 2011 at 3:29

Sampson · Accepted Answer · 2011-06-24 04:37:04Z

2

If I understand you correctly, ideally you would extend your class rather than trying to evaluate ad-hoc a body for your class:

class ClassA {
  public function greet( $msg ) {
    echo $msg;
  }
}

class ClassB extends ClassA {
  public function wave( $times ) {
    echo sprintf("Waves %s times!", $times);
  }
}

$instance = new ClassB();
$instance->greet("Hello World");
$instance->wave("five");

Note here how I'm able to build off of my class, adding new functionality and properties to the 'body' of it.

answered Jun 24, 2011 at 4:37

Sampson

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7 Comments

cwallenpoole Over a year ago

@user492983 This is the right answer. Don't dynamically create classes.

Jon Hulka Over a year ago

Not exactly. There were two problems: 1. load the class body from a second file. I've had the project requirements changed so this isn't necessary any more. 2. Load a class whose name isn't known until run-time. The class to be loaded extends a known class. Is there a way to instantiate this without using eval? What I'm doing now is eval('$instance=new '.$name.'();');

Sampson Over a year ago

@user492983 You can split the class declarations up into multiple files if needed. Just be sure to require any class you need to extend.

Jon Hulka Over a year ago

Yes, but can the class declaration be separated from its body, ie. class x extends y{<class body here>} - can <class body here> be loaded from a second file?

Sampson Over a year ago

@Jon That's a very unorthodox approach, and I cannot think of any reason outside of curiosity to take it. Did you have a good reason to do it this way?

|

user710818 · Accepted Answer · 2011-06-24 04:45:40Z

0

May by problem in non using short opening tag?

answered Jun 24, 2011 at 4:45

user710818

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