I want to pass an array in postman in which there is a image field.
But I am not getting it in to controller.
How Can I do it ?
I am passing the value like as below image.
In Header :
In my controller :
if($request->get('documentObject'))
{
foreach ($request->get('documentObject') as $documentDetails) {
if(Input::file($documentDetails['documentimage']))
{ // i am not getting it
}
How can I get it?
Make sure your form markup contains enctype multipart/form-data. Otherwise, file will be null.
For troubleshooting, you could also dump out the contends of the request to see what the browser is sending: dd($request->all()).
Related
I have a page where you can create your workout plan. Second page contains "pre-saved" workouts and I want them to load by passing parameters from second page to first. If you directly access first page, you create your workout plan from scratch.
// first page = https://prnt.sc/y4q77z
// second page = https://prnt.sc/y4qfem ; where you can check which one you want to pass to first page
// final step looks like this: https://prnt.sc/y4qh2q - but my URL looks like this:
www.example.com/training/plan?sabloni%5B%5D=84&sabloni%5B%5D=85&sabloni%5B%5D=86
this 84,85,86 are IDS
Can I pass params without changing URL ? Like having only /training/plan without anything after ?
public function plan(Request $request){
$workout = false;
if($request->workout){
$workout = $request->workout;
$workout = SablonTrening::find($sabloni); // $workout = array [1,3,4,5,6]
}
return view('trener.dodaj_trening', compact('workout'));
}
If you are getting to the /training/plan page with GET request, you could simply change it to POST. That way the parameters would be hidden in the URL but would be present in the request body. You would need a new post route:
Route::post('/training/plan', 'YourController#plan')->name('training.plan');
And then, in the form where you are selecting these plans, change the method on submit:
<form action="{{route('training.plan')}}">
//Your inputs
</form>
Your method should still work if your inputs stay the same.
Note: Not sure you would still keep the functionalities that you need, since I can't see all the logic you have.
If you have any questions, let me know.
To pass data from on blade to another blade.
At the end of first post before redirect()-route('myroute') add $request->session()->put('data', $mydata);
At the begining of the route 'myroute', just get back your data with $data = $request->old('data');
Im using Laravel v7 and i have a question about pagination.
So far im using 2 routes, 1rst to return a view with all rows from database, and 2nd receives an input and returns that view with the rows filtered by that input value.
But im using pagination, and on the 2nd route, when i try to go to 2nd page it gives me an error:
Symfony\Component\HttpKernel\Exception\MethodNotAllowedHttpException
The GET method is not supported for this route. Supported methods: POST.
I've tried to change my form method to GET but i need that the token doesn't appear on the page URL and beside that, when i go to 2nd page, it returns all the rows again.
Thats my code so far:
Routes:
Route::get('concessions', 'ConcessionController#index')->name('concessions.index');
Route::post('concessions/search', 'ConcessionController#search')->name('concessions.search');
Controller
class ConcessionController extends Controller
{
public function index()
{
$concessions = DB::table('concessions')->paginate(12);
return view('admin.concessions.index', compact('concessions'));
}
public function search(Request $request)
{
$name = $request->name;
$concessions = Concession::where('name', 'like', '%' . $name . '%')->paginate(12);
return view('admin.concessions.index', compact('concessions', 'name'));
}
}
Any way to do that?
Laravel pagination only works with get parameters.
You should use GET method for your search page. POST requests aren't meant for the purpose of displaying data. Why? There are many reasons, but to be short I will give you three examples :
1. When you access the first page, you get the data by GET request,
not POST request. So if you want to use POST request, you
need to access the page as POST request by sending data with
POST method.
2. With GET parameters, let's say you are on 5th page - you can
copy the link and paste it to friend and he will be able to view
the same content as you. With POST this is impossible.
3. You can not use back button with POST requests, if you manage to
get pagination to work.
POST requests are useful when you need to submit data to the server, in order to create new record, for example.
So I suggest you to change your route type to GET.
From my perspective, if you change your route code similar to below code, it will work properly with both methods of GET and POST.
Route::any('concessions/search', 'ConcessionController#search')->name('concessions.search');
I'me trying to use ajax with php, I have the follogin script in php :
<?php
// this file get the POST infor sent by a AJAX request and will return the value is succesful.
$price['name'] = "Called";
$price['Wheel'] = 75.25;
$price['Tire'] = 50.00;
echo json_encode($price);
?>
and I'm calling this code from my main page in the following way :
$.post("ajax/profileMod.php", {
'lname':lname,
'fname':fname,
'mname':mname,
'language':language,
'title':title,
'ptype':ptype,
'vip':vip,
'vreason':vreason
})
// Retreive the data from the php script
.done(function(data) {
// php code : echo json_encode(array("name"=>"Called!"));
alert(data);
}, "json");
// Stop original behavior
return false;
});
The returniong result from the alert is the following test :
{"name":"Called",Wheel":75.25,"Tire":50}
How can I change this result so I may use it in the following way in javascript EX:
alert(myresult['Name']) ; Would give me "Called".
So I basicly would like a associative array in javascript, but I read somewhere on this forum that you can't have associative array in Javascript, only object...
Please help!
Pass "json" as the last parameter to .post() to tell jQuery to parse the response as JSON.
(or, fix your server to return the correct Content-Type of application/json, and jQuery should do that automatically)
You will then get a Javascript object, allowing you to write
alert(result.name);
This a screen shot of what I get when I call my ajax request:
How do I run only the task, without printing the whole page? This is my ajax call:
$.ajax
({
type: "POST",
url: "index.php?option=com_similar&task=abc",
data: {
id: id,
name: name,
similar_id: similar_id,
},
cache: false,
success: function(html)
{
$("#flash").fadeOut("slow");
$("#content"+similar_id).html(html);
}
});
});
$(".close").click(function()
{
$("#votebox").slideUp("slow");
});
});
Don't go with exit or die, Joomla! has it's nice way of dealing with this stuff.
The answers below are tested in Joomla! 2.5 & 3 (for 1.5. may work as well).
General
Your URL for the task needs to look like this:
index.php?option=com_similar&task=abc&format=raw
You than create the controller which will use the view, let's say Abc, which will contain the file view.raw.html (identical to a normal view file).
Below you have the code for generate a raw HTML response:
/controller.php
public function abc()
{
// Set view
JRequest::setVar('view', 'Abc');
parent::display();
}
/views/abc/view.raw.php
<?php
defined('_JEXEC') or die;
jimport('joomla.application.component.view');
class SimilarViewAbc extends JView
{
function display($tpl = null)
{
parent::display($tpl);
}
}
/views/abc/tmpl/default.php
<?php
echo "Hello World from /views/abc/tmpl/default.php";
Note: This is the solution I would use if I had to return HTML (it's cleaner and follows Joomla logic). For returning simple JSON data, see below how to put everything in the controller.
If you make your Ajax request to a subcontroller, like:
index.php?option=com_similar&controller=abc&format=raw
Than your subcontroller name (for the raw view) needs to be abc.raw.php.
This means also that you will / may have 2 subcontrollers named Abc.
If you return JSON, it may make sense to use format=json and abc.json.php. In Joomla 2.5. I had some issues getting this option to work (somehow the output was corrupted), so I used raw.
If you need to generate a valid JSON response, check out the docs page Generating JSON output
// We assume that the whatver you do was a success.
$response = array("success" => true);
// You can also return something like:
$response = array("success" => false, "error"=> "Could not find ...");
// Get the document object.
$document = JFactory::getDocument();
// Set the MIME type for JSON output.
$document->setMimeEncoding('application/json');
// Change the suggested filename.
JResponse::setHeader('Content-Disposition','attachment;filename="result.json"');
echo json_encode($response);
You would generally put this code in the controller (you will call a model which will return the data you encode - a very common scenario). If you need to take it further, you can also create a JSON view (view.json.php), similar with the raw example.
Security
Now that the Ajax request is working, don't close the page yet. Read below.
Don't forget to check for request forgeries. JSession::checkToken() come in handy here. Read the documentation on How to add CSRF anti-spoofing to forms
Multilingual sites
It may happen that if you don't send the language name in the request, Joomla won't translate the language strings you want.
Consider appending somehow the lang param to your request (like &lang=de).
New in Joomla 3.2! - Joomla! Ajax Interface
Joomla now provides a lightweight way to handle Ajax request in a plugin or module. You may want to use the Joomla! Ajax Interface if you don't have already a component or if you need to make requests from a module your already have.
If you just want to include the response output in some HTML element, append format=raw to your URL as mentioned above. Then you could have a controller function like this:
function abc(){
//... handle the request, read variables, whatever
print "this is what I want to place in my html";
}
The AJAX response will output everything you printed / echoed in the controller.
how can i get folowing url
http://localhost:4847/Category/#pageindex=1
i failed to get full url using Request.Url.OriginalString
its only give me blocalhost:4847/Category
ignore other paramters. basically i want to get #pageindex=1 from this url
there is any other way to get (#pageindex=1)
Why are you using # ? Is it hardcoded ? Why not using Html helper methods to generate the Link urls ?
#Html.ActionLink("Category", "Index", new { pageIndex= "1"})
your query string should be like
http://localhost:4847/Category?pageindex=1
Your request for this page will be handled by an Action. So i guess, you should be able to get this as the parameter of that action method.
public ActionResult Index(string pageIndex)
{
//do whatver with pageIndex variable value here
}