How do I pass all form data function to third party URL. Suppose my third URL is
`http:\\www.abletoaccess.com\form\request`
For the security reason I don't want to access this URL in form action method or I don't want to post direct form data. I want when I submit the form all data comes in my function and redirect to third party URL with post data and added more parameters.
Any help will be appreciated!!!!
You can handle the request by your controller and then show view and auto-submit it to external url.
<div class="container">
<div class="row">
<div class="col-md-8 col-md-offset-2">
<div class="panel panel-default">
<div class="panel-heading">
Page is loading...
</div>
<div class="panel-body">
{!! Form::open(['url' => $externalUrl, 'method' => 'POST', 'class' => 'form-horizontal', 'id' => 'my-form']) !!}
@foreach ($fields as $key => $value)
{{ Form::hidden($key, $value) }}
@endforeach
{{ Form::hidden('signature', $signature) }}
{!! Form::close() !!}
</div>
</div>
</div>
</div>
</div>
<script>
document.getElementById('my-form').submit();
</script>
Try Following Code:
<form method="post" action={{ action('Controller@method') }}>
<input type="submit" value="add">
</form>
in controller file write below code :
public function method(Request $request)
{
Redirect::away('external url')->withInputs(Input::all());
}
Laravel 5: how to redirect with data to external resource form controller
First of all, this seems not to be Laravel specific issue.
Second, you need to check CURL function in PHP.
If you can, install Guzzle client that helps with these things.
You can pass the client to your controller and do a post request to another URL like this:
public function someMethod(GuzzleHttp\Client $client)
{
$client->post('http:\\www.abletoaccess.com\form\request', [
'form_params' => [
'param1' => 'something',
...
]
]);
}
