How to use var_dump in joomla - php

My question might be little silly but please bear with me. I suppose var_dump should work anywhere in the code which calls its service but unfortunately i can't return anything if i use it in a controller, or model. ya it does work in the view/layout page.
I tried testing the following simple thing in one of my controller function and it returns nothing;
$foo = "bar";
var_dump($foo);
Please enlighten me!

I don't know Joomla, but in an MVC framework, the view expects data to come from the controller in a particular format, perhaps JSON or XML. When you call var_dump(), it will most likely mess up the syntax of this, so the application won't work. When you're debugging with this tool, you'll want to make use of the browser's console (Developer Tools or Firebug) to view what was returned. Go into the Network tab, select the URL of the controller, and then view the response data. There you'll see the output of var_dump().

For show a variable and pause the executing you have 3 options.
echo $variable;
print_r($variable);
var_dump($variable);
and you need to write die() after them to stop code and show your $variable.

Related

How to print query result in python/django

I came from CakePHP and just started playing with Django framework. In CakePHP, I have the habit of printing out all the returned array using pr() straight out to the webpage. For example:
A controller spits out a $result to a View, I use pr($result) and it will print out everything right on the webpage so I know how to travel through $result from my View.
A form POST a $request to a Controller, I use pr($request) to see what is sending in before processing it in the Controller. The content of $request will be displayed immediately on the webpage right after I hit Submit the form.
I'm wondering if I could do the same thing in django instead of going to the shell and try pprint (or could I just use pprint to print out to the web???)
This is a really simple example about what I'm talking about:
app_name/views.py:
def detail(request, soc_id):
soc = get_object_or_404(Soc, pk=soc_id)
return render(request, 'socs/detail.html', {'soc': soc})
How can I just view clearly what is in "soc". In cakephp, I could just pr($soc) right there and it will be displayed right on the detail.html page.
I have tried this and it didn't work (I'm sure it's basic but i'm just new to this)
import pprint
def detail(request, soc_id):
soc = get_object_or_404(Soc, pk=soc_id)
pprint.pprint(soc)
return render(request, 'socs/detail.html', {'soc': soc})
I have spent two days doing research but I couldn't find the answer. I'm hoping one of you can help this newbie out.
The way you're trying to print will show the print in the terminal that is running your Django server. If you just need to see it quick, check there.
If you want to output the value on to the rendered page, you'll have to include it in the template using template tages. It looks like you send {'soc': soc} to your template as context. Because of this, you should be able to use it in your template. So, in your template (socs/detail.html), just add {{ soc }} somewhere and it should print out the value. The template has full access to the object, so if you wanted something specific, you can also print that instead (like {{ soc.id }}).
If you want to see everything that's in the object without specifying all of the different fields yourself, send OBJECT.__dir__. For example, for this you'd send {'soc': soc.__dir__} as your context. Keep in mind that this likely should not be used for anything but inspection on your part.
If you'd like to learn more about Django's templating, check out the syntax and more.

PHP Error handler triggers when using die in outputbuffer context

im developing my PHP framework and at the moment im stuck with a strange behaviour in my error-dispatching page.
The page itself is beautiful, it catches and lists all error using a custom error-handler registered with "set_error_handler()" and can print the line, file, invoked variables and such. It even triggers multiple errors perfectly (except for fatal or syntax-errors which causes the script to hold) but still the dispatches the fatal or even syntax error in this case and even shows me, even when its a null-pointer in a template where the error is located which is beautiful and works awesome.
This system works "using set_error_handler"! The entire output (rendering html /php templates) is realised using "ob_start" and "ob_get_clean" - outputbuffer context.
Next I check if the error-handler catched any errors (will be written in an array in the error-handler's class) and if there are errors and error-display is enabled (developer mode = on) the error-dispatcher page will be displayed instead of the page's content.
However since I implemented that I have one strange behaviour which I do not want:
Whenever I use die() inside of outputbuffer context ((ob_start() ... ob_get_clean()) ) The registered error-handler method is called with an empty error-list displayed and even error_get_last() is empty, but still he triggers that damn function.
This is very disturbing as you cant use die to hold the script to test the output of certain variables which is disturbing in development and seems to be an unusual behaviour to me.
I think posting my entire code is making the thing a little bit too complicated for you to understand thatswhy I leave a little draft explaining the situation, its basically same as it looks in my classes:
//index.php
$errorList = array();
function myErrorHandler($errorType, $errorString, $errorFile, $errorLine, array $errorContext) {
global $errorList;
$errorList[] = $errorString;
}
set_error_handler("myErrorHandler");
//Here my classses and entire rest of framework is loaded...()
//Initialise output
$output = "";
//This method parses templates, it remotely close works like that...
function includeTemplate($templatePath) {
global $output;
ob_start("myErrorHandler");
include($templatePath);
$output .= ob_get_clean(); //Attach to output
}
//This method is called at the very end, it flushes the output, wether by displaying error-dispatcher or page content to be shown.
function flush() {
global $output;
if(count($errorList) > 0) {
echo implode("<br />", $errorList);
} else {
echo $output;
}
}
//Template file: registerTemplate.php
<div class="container">
<Some template output here, doesnt matter if an error occured...>
<?php die($var); ?> //<- Error rised -> Entering "myErrorHandler()" -> but why????
</div>
-> This die() above would rise an error because registerTemplate.php was somewhen parsed using includeTemplate("reigsterTemplate"); which is as located within an ob_start() context. It does not matter if an error occured before or not or even if I just place die("!"); without any invoked variable, it will riase an error.
If I put the die() before or after the include_template context, in upper scope it works as usual and the error_handler does not trigger, as it should work.
Can anyone tell me why this is happening? Thanks alot!
(Important: again this code is not realistic I know its just a draft from how the things look like very barely, dont bother with that giving me hints that this is a bad way to do things. The entire thing is perfectly object oriented and very well structured. I just made this little draft to make you guys understand in what order the methods are called and how my problem comes up because posting my code would waste me and you alot of time as it would be too huge and complex here).

Better way to output PHP objects to console?

I'm trying to find the best way to debug PHP in the browser.
EDIT:
I know about xdebug, var_dump and print_r; that's not the point. a) I want a way to debug when a site is live (I know, that's bad but sometimes have to for a quick fix), b) dumping PHP variables all over a web page is ugly, and c) I sometimes need to know why AJAX calls aren't working, and it's easier to look at JSON in the console.
Possible other solutions include wrapping your debug output in a hidden div, e.g.
<div id="debug"><?php var_dump($all_the_junk) ?></div><script>document.getElementById('debug').style.display='none'</script>
and then typing something like
document.getElementById('debug')style.display = 'block'
into the console.
But let's be honest, that kind of sucks, and defeats the purpose of being able to see everything in a tidy way.
ENDEDIT;
So, anyways, I have the following code:
function log_as_json($vars)
{
if(is_resource($vars)) {
return;
}
else {
if($vars)
{
$json = #json_encode($vars);
print "<script>console.log($json);</script>";
}
}
}
The problem is that once in a while I get these terrible recursions, but I'm not sure what the best way is to check for types that json_encode supports. Also I suspect there is a better way to do this. Advice would be appreciated. Thanks!
If you are really keen on debugging in the browser, I'd recommend Chrome Logger (formerly known as ChromePHP).
The best way is to use xdebug And an IDE that allows you to set breakpoints
I use print_r or var_dump - it is what they are designed for.
I suppose you could embed these calls in console.log - not sure how they would display though.
I know this is old but here is a way that works for me, just call the function when you need to..
function console_log( $data ){
echo '<script>';
echo 'console.log('. json_encode( $data ) .')';
echo '</script>';
}

mysql fetch array returns false in function but not in the document

i am creating a CMS and have php creating a page. i have a while loop like this
while($row = mysql_fetch_array($results)) {
echo "some html code" . $row['name'];
its shortend but hopefully you get the point. i have the full thing in my page working just as it should and i wanted to move it to a function include as i want to reuse it. the problem is i do that and it stops working.
i did some testing and found that the function is getting the query result and after doing a var dump both were identical the problem comes when i try to assign it to an array. it comes back as false so in the above code, for example,
$row = false;
im toatly lost in this and if my explanation is confusing i appologise but i am a bit of a newbie i have tried searching but....i dont really know where to begin
any thoughts.
the query you are doing is basically wrong, try posting exactly the code which you have in $query and then let us see the problem.
also, it is better to use mysqli functions.
but for this, edit the question and type the query, or simply put a die(mysql_error()) at the end of your query which is in $query. It will show your exact error.
i fugured it out
when i was testing the function i commented out the original code on the main page but for some reason i had not comented out enough (it was a mix of php and html clearly the php had not been commented out properly) this must have been causing a clash of some kind as when i put the function above the code on my page the function worked and the long code below it did not
sorry for wasting your time guys

how do I know the reason of saving failure in CakePHP?

I am new to cakePHP.t I thought this should be simple but I am having hard time to find the reason why saving in a controller action does fail.
I tried to print error
if ($this->MembershipRequest->save($this->data['MembershipRequest'])){
$this->flash("All items are saved", array("controller"=>"home", "action"=>"index"));
} else {
pr($this->MembershipRequest->validationErrors);
}
But it does not print anything. Could anybody give me right direction?
I am looking for a good way of debugging cakephp app. One thing I just found out is Configure::write('debug',...) which gives great debugging messages. Is there any other tool? Thanks
I'm not sure of your format of $this->data, but I'm pretty sure you don't need to pass the "MembershipRequest' as a key in the save, try saving with $this->MembershipRequest->save($this->data) and it should work.
Also, check the beforeSave() methods you have in your MembershipRequest or AppModel, they should always return true, otherwise the save will silently fail.
For other debugging help, I suggest you look at the CakePHP DebugKit.

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