1
votes

my application.ini

[production]
phpSettings.display_startup_errors = 0
phpSettings.display_errors = 0
includePaths.library = APPLICATION_PATH "/../library"

bootstrap.path = APPLICATION_PATH "/Bootstrap.php"
bootstrap.class = "Bootstrap"
test.bootstrap.path = APPLICATION_PATH "/modules/test/Bootstrap.php"
test.bootstrap.class = "Test_Bootstrap"

appnamespace = "Application"

resources.frontController.controllerDirectory = APPLICATION_PATH "/controllers"
resources.frontController.params.displayExceptions = 0
resources.layout.layoutPath = APPLICATION_PATH "/layouts/scripts/"

resources.view.basePath = APPLICATION_PATH "/views/"

resources.view[] =
test.resources.view[] = 

db.adapter = "PDO_MYSQL"
db.params.dbname = "money"
db.params.username = "root"
db.params.password = "**************"

resources.modules[] =
resources.frontController.moduleDirectory = APPLICATION_PATH "/modules"
[staging : production]

[testing : production]
phpSettings.display_startup_errors = 1
phpSettings.display_errors = 1

[development : production]
phpSettings.display_startup_errors = 1
phpSettings.display_errors = 1
resources.frontController.params.displayExceptions = 1

application/modules/test/bootstrap.php

class Test_Bootstrap extends Zend_Application_Module_Bootstrap
{
    protected function _initLeftMenu()
    {
        $this->bootstrap('View');
        $view = $this->getResource('View');
        $view->render('index/_left_menu.phtml'); // <-- here error
    }
}

have a problem with $view->render

Fatal error: Uncaught exception 'Zend_View_Exception' with message 'no view script directory set; unable to determine location for view script' in C:\ZendFramework\library\Zend\View\Abstract.php:973 Stack trace: #0 C:\ZendFramework\library\Zend\View\Abstract.php(884): Zend_View_Abstract->_script('_left_menu.phtm...') #1 E:\www\money2\application\modules\test\Bootstrap.php(19): Zend_View_Abstract->render('_left_menu.phtm...') #2 C:\ZendFramework\library\Zend\Application\Bootstrap\BootstrapAbstract.php(667): Test_Bootstrap->_initLeftMenu() #3 C:\ZendFramework\library\Zend\Application\Bootstrap\BootstrapAbstract.php(620): Zend_Application_Bootstrap_BootstrapAbstract->_executeResource('leftmenu') #4 C:\ZendFramework\library\Zend\Application\Bootstrap\BootstrapAbstract.php(584): Zend_Application_Bootstrap_BootstrapAbstract->_bootstrap(NULL) #5 C:\ZendFramework\library\Zend\Application\Resource\Modules.php(124): Zend_Application_Bootstrap_BootstrapAbstract->bootstrap() #6 C:\ZendFramework\library\Zend\Application\Resource\Mod in C:\ZendFramework\library\Zend\View\Abstract.php on line 973

any idea?

2
Why don't you start sentences with upper case characters? - takeshin
Somebody needs to edit this question later to meet the SO standards. This is a real people community, other folks are reading your posts over and over. You could consider some additional micro time to write the question properly as we are taking the time to figure out the answer for your question. - takeshin

2 Answers

0
votes

find own salvation: application.ini

test.resources.view.basePath = APPLICATION_PATH "/modules/test/views/"
0
votes

remove test.resources.view[] =

View resource in module replaces view from main bootstrap.

Hint:
$this->getApplication(); in module bootstrap will return main bootstrap
Use it to bootstrap and retrieve view.

class Test_Bootstrap extends Zend_Application_Module_Bootstrap
{
    protected function _initLeftMenu()
    {
        //main bootstrap, injected by modules resource
        $bootstrap = $this->getApplication();
        $bootstrap->bootstrap('View');
        $view = $bootstrap->getResource('View');
        $view->render('index/_left_menu.phtml'); // <-- here error
    }
}

btw, bootstrap isn't a good place to render something. Consider moving it to more appropriate places. Action helper for example.
Also check this blog post