I'm using Laravel 5.8, and have several input fields which of course has an old() directive on every value="" tag.
This is my example right now:
<input class="form-control input-md" name="contact_name" type="text" value="#if($edit){{ $ad->contact_name }}#else{{ old('contact_name')}}#endif">
I now that if I use this: {{ old('contact_name', "John")}}
The default value will be "John"
But I want to do a check if there is a user logged in and prefill that input with the User contact name.
My idea is something like this:
value="#if($edit){{ $ad->contact_name }}#else{{ old('contact_name', Auth::user()->name)}}#endif
And it works! But of course, it throws: Trying to get property 'name' when I get an incognito window.
So, how do I evaluate logged in users and prefill this?
You can use the optional helper:
{{ old('contact_name', optional(Auth::user())->name) }}
Related
I have the following URL:
https://example.com/?email=test#test.com
then I have the following blade template
<input value="{{Request::query('email') or old('email')}}">
But it's not displaying the email I passed in the get parameter into the input. Instead, it's displaying the value 1
Tried searching haven't found a solution that works.
That's because you have {{Request::query('email') or old('email')}} which is conditional. It will always return true or false.
Try any of these:
{{Request::query('email') || old('email')}}
or
{{Request::query('email') ? Request::query('email') : old('email')}}
You can try this:
<input value="{{ request()->email }}">
Having issue with form validation .
i want to submit the form only when form is valid.
but with the empty inputs and clicking on submit button is submitting the form although the inputs are empty.
<form name="equipmentForm" #f="ngForm" (ngSubmit)="f.form.valid && addEquipment()" validate>
Inputs be like this.
<input name="equimentId" class="text-input form-control" type="text" [(ngModel)]="model.equipmentNumber" pattern="^[0-9][0-9]{1,19}$" title="Equipment ID. can be upto 20 digits only.">
I cant post the whole code although.
this
f.form.valid is true from form initialization
wanted to acheive something like this
<div *ngIf="!model.equipmentModel && f.submitted" class="text-danger">
Please enter Equipment Model
</div>
So on submit i want to show this message instead of default browser's.
but this f.form.valid is goddamn true from default.
You should add required attribute to your input tags to, then as #Cobus Kruger mentioned, form will not be submitted untill it is filled.
However you can also give a try to pristine, dirty options, which allow you to check if the user did any changes to the form so in this case your condition may look like this:
<form name="equipmentForm" #f="ngForm" (ngSubmit)="f.form.valid && f.form.dirty ? addEquipment() : ''" validate>
and the input:
<input name="equimentId" class="text-input form-control" type="text" [(ngModel)]="model.equipmentNumber" pattern="^[0-9][0-9]{1,19}$" title="Equipment ID. can be upto 20 digits only." required />
In this case it will check if any changes were applied to the input, and submit the form if both conditions are met.
If you specify the required attribute on the input, then the form will not be submitted unless a value is filled in. But that only covers values that were not supplied and you may want to check for invalid values as well.
The usual way is to disable the submit button unless the form is valid. Like this:
<button type="submit" [disabled]="!f.form.valid">Submit</button>
The Angular documentation about form validation also shows this. Look near the bottom of the "Simple template driven forms" section
In function which you call on submit you can pass form as parameter and then check. In html you will need to pass form instance:
<form name="equipmentForm" #f="ngForm" (ngSubmit)="addEquipment(f)" validate>
In typescript:
addEquipment(form){
if(form.invalid){
return;
}
//If it is valid it will continue to here...
}
if the validation fails it's not clear to me how i could pass the old input to fill again the form.
I mean, I know how to pass the data when using the Validator class and redirecting after fail with the method withInput(), but I'm trying to learn how to use Form Requests provided in laravel 5. Thanks
$username = Request::old('username');
or in view:
{{ old('username') }}
Read more: http://laravel.com/docs/5.0/requests#old-input
You can redirect with old data like below with validation error
return redirect()->back()->withErrors($validator)->withInput();
<input type="text" name="username" value="{{ old('username') }}">
you must also define withInput() for the http route when redirecting to a page for example
return back()->withInput();
Im making a edit form for my app and i was wondering if someone could tell me how to get the data from the database into my text field.
I can locate the record i need to edit based on the users click, and i can display the information if i do the following:
value="{{ $letter->subject }}"
BUT, the problem im having is that when i run it through the validation and there is an error, it comes back with the database information instead of the OLD data.
So my questions is. Is there a way to serve up the database information first and then when it goes through the validatior, validate the information the user has edited?
Currently to validate the text field and bring the data back incase of error, im using
Input::old('subject')
Is there a parameter for that old bit that allows me to put in the DB data?
Cheers,
Hey you could validate and return ->withInput() and then in your actual form, check if there is Input::old() and display it, otherwise display from the db.
example:
<input type="text" name="subject"
value="{{ (Input::old('subject')) ? Input::old('subject') : $letter->subject }}">
Or you could go the other way and define the variable and do a regular if statement, instead of the ternary one! Up to you to decide what you want to use!
All you need is form model binding http://laravel.com/docs/html#form-model-binding:
{{ Form::model($letter, ['route' => ['letters.update', $letter->id], 'method' => 'put']) }}
// your fields like:
{{ Form::text('someName', null, ['class' => 'someHTMLclass' ...]) }}
// no default values like Input::old or $letter->something!
{{ Form::close() }}
This way you form will be populated by the $letter data (passed from the controller for example).
Now, if you have on your countroller:
// in case of invalid data
return Redirect::back()->withInput();
then on the redirect your form will be repopulated with input values first, not the original model data.
Make it more simple and clean
<input type="text" name="subject" value="{{ (Input::old('subject')) ?: $letter->subject }}">
I'm not sure for Laravel 4 but in Laravel 5, function old takes second param, default value if no old data in session.
Please check this answer Best practice to show old value
I know browsers only support POST and GET requests, and Laravel supports PUT requests using the following code:
<?= Form::open('/path/', 'PUT'); ?>
... form stuff ...
<?= Form::close(); ?>
This produces the following HTML
<form method="POST" action="http://example.com/home/" accept-charset="UTF-8">
<input type="hidden" name="_method" value="PUT" />
... form stuff ...
</form>
How does the framework handle this? Does it capture the POST request before deciding which route to send the request off to? Does it use ajax to send an actual PUT to the framework?
It inserts a hidden field, and that field mentions it is a PUT or DELETE request
See here:
echo Form::open('user/profile', 'PUT');
results in:
<input type="hidden" name="_method" value="PUT">
Then it looks for _method when routing in the request.php core file (look for 'spoofing' in the code) - and if it detects it - will use that value to route to the correct restful controller.
It is still using "POST" to achieve this. There is no ajax used.
Laravel uses the symfony Http Foundation which checks for this _method variable and changes the request to either PUT or DELETE based on its contents. Yes, this happens before routing takes place.
You can also use an array within your form open like so:
{{ Form::open( array('route' => array('equipment.update', $item->id ),
'role' => 'form',
'method' => 'put')) }}
Simply change the method to what you want.
While a late answer, I feel it is important to add this for anyone else who finds this and can't get their API to work.
When using Laravel's resource routes like this:
Route::resource('myRoute','MyController');
It will expect a PUT in order to call the update() method. For this to work normally (outside of a form submission), you need to make sure you pass the ContentType as x-www-form-urlencoded. This is default for forms, but making requests with cURL or using a tool like Postman will not work unless you set this.
PUT usually refers to update request.
When you open a form inside laravel blade template using,
{{ Form::open('/path/', 'PUT') }}
It would create a hidden field inside the form as follows,
<input type="hidden" name="_method" value="PUT" />
In order for you to process the PUT request inside your controller, you would need to create a method with a put prefix,
for example, putMethodName()
so if you specify,
{{ Form::open('controller/methodName/', 'PUT') }}
inside Form:open. Then you would need to create a controller method as follows,
class Controller extends BaseController {
public function putMethodName()
{
// put - usual update code logic goes here
}
}