Re: Cake v2.x -- Looking to return rendered view to the Controller
Just looking at the API code I know that'll work. Ugh. Looking through all
of the files yesterday I'm surprised I missed it. Thanks, Andrew!!!
On Tuesday, February 10, 2015 at 11:22:28 PM UTC-5, Andrew Lechowicz wrote:
My bad. In typical Cake fashion, a method with the name of the property
you want to get when passed null returns that property. So my code above
should be
$emailBody = $this-response-body();
Note the parentheses. For your reference:
http://api.cakephp.org/2.6/source-class-CakeResponse.html#605-617
On Tuesday, February 10, 2015 at 7:44:55 PM UTC-5, BrendonKoz wrote:
Thank you for the response, Andrew. In my test (using a view file that
just contains static text), $this-response-body seemed to return an empty
string due to ViewBlock-get having an empty $ViewBlock-_blocks value (as
to what $_blocks is or how it's set, I've no idea).
I don't suppose anyone has any other ideas?
I've tried switching to using CakeEmail for sending email to take
advantage of layouts and views and there's a bug preventing me from
connecting to my SMTP server's TLS connection (which I have to report), so
neither solution is currently working for me. :-/ Getting templates to work
with a third party library would be easier for me!
On Thursday, February 5, 2015 at 4:14:48 PM UTC-5, Andrew Lechowicz wrote:
It looks like `Controller::render()` sets the body on the CakeResponse
object and then returns the CakeResponse object:
http://api.cakephp.org/2.6/source-class-Controller.html#922-962. I
would imagine you could access the rendered view like so:
public function test()
{
$this-render('/Emails/html/test', false);
$emailBody = $this-response-body;
// Do what you want with $emailBody here
}
On Wednesday, February 4, 2015 at 5:33:35 PM UTC-5, BrendonKoz wrote:
For various reasons I had decided to use a 3rd party email library
within my current CakePHP project. I thought that it might be nice to use
CakePHP's Views to create templating for my emails and take advantage of
layouts too. Unfortunately I'm stumped on just how to retrieve the
rendered
output of a view back to the Controller method.
I've tried the following:
public function test() {
$this-layout = false;
#$this-view = '/Emails/html/test.ctp';
$var = $this-render('/Emails/html/test', false);
pr($var-_body); die();
}
$var-_body is protected (as denoted by the underscore). I saw no other
property within the $var variable that contained the body code within my
(template) view.
Are there any ways to do this that I'm just not seeing? If so, can I
safely presume that layouts would be handled in a similar fashion?
Thank you for any possible help...
--
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Re: Cake v2.x -- Looking to return rendered view to the Controller
My bad. In typical Cake fashion, a method with the name of the property you
want to get when passed null returns that property. So my code above
should be
$emailBody = $this-response-body();
Note the parentheses. For your reference:
http://api.cakephp.org/2.6/source-class-CakeResponse.html#605-617
On Tuesday, February 10, 2015 at 7:44:55 PM UTC-5, BrendonKoz wrote:
Thank you for the response, Andrew. In my test (using a view file that
just contains static text), $this-response-body seemed to return an empty
string due to ViewBlock-get having an empty $ViewBlock-_blocks value (as
to what $_blocks is or how it's set, I've no idea).
I don't suppose anyone has any other ideas?
I've tried switching to using CakeEmail for sending email to take
advantage of layouts and views and there's a bug preventing me from
connecting to my SMTP server's TLS connection (which I have to report), so
neither solution is currently working for me. :-/ Getting templates to work
with a third party library would be easier for me!
On Thursday, February 5, 2015 at 4:14:48 PM UTC-5, Andrew Lechowicz wrote:
It looks like `Controller::render()` sets the body on the CakeResponse
object and then returns the CakeResponse object:
http://api.cakephp.org/2.6/source-class-Controller.html#922-962. I would
imagine you could access the rendered view like so:
public function test()
{
$this-render('/Emails/html/test', false);
$emailBody = $this-response-body;
// Do what you want with $emailBody here
}
On Wednesday, February 4, 2015 at 5:33:35 PM UTC-5, BrendonKoz wrote:
For various reasons I had decided to use a 3rd party email library
within my current CakePHP project. I thought that it might be nice to use
CakePHP's Views to create templating for my emails and take advantage of
layouts too. Unfortunately I'm stumped on just how to retrieve the rendered
output of a view back to the Controller method.
I've tried the following:
public function test() {
$this-layout = false;
#$this-view = '/Emails/html/test.ctp';
$var = $this-render('/Emails/html/test', false);
pr($var-_body); die();
}
$var-_body is protected (as denoted by the underscore). I saw no other
property within the $var variable that contained the body code within my
(template) view.
Are there any ways to do this that I'm just not seeing? If so, can I
safely presume that layouts would be handled in a similar fashion?
Thank you for any possible help...
--
Like Us on FaceBook https://www.facebook.com/CakePHP
Find us on Twitter http://twitter.com/CakePHP
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Re: Cake v2.x -- Looking to return rendered view to the Controller
Thank you for the response, Andrew. In my test (using a view file that just
contains static text), $this-response-body seemed to return an empty
string due to ViewBlock-get having an empty $ViewBlock-_blocks value (as
to what $_blocks is or how it's set, I've no idea).
I don't suppose anyone has any other ideas?
I've tried switching to using CakeEmail for sending email to take advantage
of layouts and views and there's a bug preventing me from connecting to my
SMTP server's TLS connection (which I have to report), so neither solution
is currently working for me. :-/ Getting templates to work with a third
party library would be easier for me!
On Thursday, February 5, 2015 at 4:14:48 PM UTC-5, Andrew Lechowicz wrote:
It looks like `Controller::render()` sets the body on the CakeResponse
object and then returns the CakeResponse object:
http://api.cakephp.org/2.6/source-class-Controller.html#922-962. I would
imagine you could access the rendered view like so:
public function test()
{
$this-render('/Emails/html/test', false);
$emailBody = $this-response-body;
// Do what you want with $emailBody here
}
On Wednesday, February 4, 2015 at 5:33:35 PM UTC-5, BrendonKoz wrote:
For various reasons I had decided to use a 3rd party email library within
my current CakePHP project. I thought that it might be nice to use
CakePHP's Views to create templating for my emails and take advantage of
layouts too. Unfortunately I'm stumped on just how to retrieve the rendered
output of a view back to the Controller method.
I've tried the following:
public function test() {
$this-layout = false;
#$this-view = '/Emails/html/test.ctp';
$var = $this-render('/Emails/html/test', false);
pr($var-_body); die();
}
$var-_body is protected (as denoted by the underscore). I saw no other
property within the $var variable that contained the body code within my
(template) view.
Are there any ways to do this that I'm just not seeing? If so, can I
safely presume that layouts would be handled in a similar fashion?
Thank you for any possible help...
--
Like Us on FaceBook https://www.facebook.com/CakePHP
Find us on Twitter http://twitter.com/CakePHP
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Re: Cake v2.x -- Looking to return rendered view to the Controller
It looks like `Controller::render()` sets the body on the CakeResponse
object and then returns the CakeResponse
object: http://api.cakephp.org/2.6/source-class-Controller.html#922-962. I
would imagine you could access the rendered view like so:
public function test()
{
$this-render('/Emails/html/test', false);
$emailBody = $this-response-body;
// Do what you want with $emailBody here
}
On Wednesday, February 4, 2015 at 5:33:35 PM UTC-5, BrendonKoz wrote:
For various reasons I had decided to use a 3rd party email library within
my current CakePHP project. I thought that it might be nice to use
CakePHP's Views to create templating for my emails and take advantage of
layouts too. Unfortunately I'm stumped on just how to retrieve the rendered
output of a view back to the Controller method.
I've tried the following:
public function test() {
$this-layout = false;
#$this-view = '/Emails/html/test.ctp';
$var = $this-render('/Emails/html/test', false);
pr($var-_body); die();
}
$var-_body is protected (as denoted by the underscore). I saw no other
property within the $var variable that contained the body code within my
(template) view.
Are there any ways to do this that I'm just not seeing? If so, can I
safely presume that layouts would be handled in a similar fashion?
Thank you for any possible help...
--
Like Us on FaceBook https://www.facebook.com/CakePHP
Find us on Twitter http://twitter.com/CakePHP
---
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to cake-php+unsubscr...@googlegroups.com.
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Cake v2.x -- Looking to return rendered view to the Controller
For various reasons I had decided to use a 3rd party email library within
my current CakePHP project. I thought that it might be nice to use
CakePHP's Views to create templating for my emails and take advantage of
layouts too. Unfortunately I'm stumped on just how to retrieve the rendered
output of a view back to the Controller method.
I've tried the following:
public function test() {
$this-layout = false;
#$this-view = '/Emails/html/test.ctp';
$var = $this-render('/Emails/html/test', false);
pr($var-_body); die();
}
$var-_body is protected (as denoted by the underscore). I saw no other
property within the $var variable that contained the body code within my
(template) view.
Are there any ways to do this that I'm just not seeing? If so, can I safely
presume that layouts would be handled in a similar fashion?
Thank you for any possible help...
--
Like Us on FaceBook https://www.facebook.com/CakePHP
Find us on Twitter http://twitter.com/CakePHP
---
You received this message because you are subscribed to the Google Groups
CakePHP group.
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to cake-php+unsubscr...@googlegroups.com.
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Visit this group at http://groups.google.com/group/cake-php.
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