note 52187 added to function.var-export

Tip for using the _var_export() code below with objects (top level, or nested): In the original design of the function, the same class name is used. This, however, may cause a conflict with the original class definition. To make sure all exports are created without conflicting with the original class definition, change this line... <? if ($class_name == "stdClass") $class_name = "_{$class_name}_"; ?> to this... <? $class_name = "_{$class_name}_"; ?> Regards. ---- Manual Page -- http://www.php.net/manual/en/function.var-export.php Edit -- http://master.php.net/manage/user-notes.php?action=edit+52187 Delete: added to the manual -- http://master.php.net/manage/user-notes.php?action=delete+52187&report=yes&reason=added+to+the+manual Delete: bad code -- http://master.php.net/manage/user-notes.php?action=delete+52187&report=yes&reason=bad+code Delete: spam -- http://master.php.net/manage/user-notes.php?action=delete+52187&report=yes&reason=spam Delete: useless -- http://master.php.net/manage/user-notes.php?action=delete+52187&report=yes&reason=useless Delete: other reasons -- http://master.php.net/manage/user-notes.php?action=delete+52187&report=yes Reject -- http://master.php.net/manage/user-notes.php?action=reject+52187&report=yes Search -- http://master.php.net/manage/user-notes.php