Codeigniter 3
I am using validation form using required and a callback function.
But it doesn't work, I am in edit controller and view.
When I call the callback function with a cellular that does exist it tells me the cellular exists. But when I try to change the cellular to another it tells me
You must enter the cellular, Just cannot change the cellular number to another new one.
<php echo form_error('volasoccel' );>
<input value='$volasoc[0]->Celular; required type='text' name="volasoccel" class="form-control" id="volasoccel" placeholder="Celular" >
$this->form_validation->set_rules('volasoccel', 'Celular',required| 'callback_volasoccel_check[tvolasoc.TVOLASOC_CEL]|numeric|trim|xss_clean',
array('required' => 'Debe ingresar el Celular'));
Your required should be enclosed with a single quote:
'required| callback_volasoccel_check[tvolasoc.TVOLASOC_CEL]|numeric|trim|xss_clean'
Related
I work with playFramework 2.6. I have a popup that contains a form. When I try to submit the form, I get the following error:
Erreur Client!403 - No CSRF token found in body.
How can I fix this error?
Enable CSRF folder within application.conf:
play.filters.enabled += "play.filters.csrf.CSRFFilter"
Then in your form have this:
<form action="/submit-form-url" method="POST">
#CSRF.formField
//Your form body
</form>
In case you dont want to use the helper (#CSRF.formField)
First import the required library: import play.filters.csrf.CSRF.
Second, get its value within the controller method: CSRF.getToken and pass it on the views.
Third, within the views (after the form initial line) use the hidden input to hold the token value:
<input type="hidden" name="csrfToken" value="whatever-that-is">
Does anybody know how to validate the file type input.
I have modified (hard coded) the class.vf_file.php input.
$strOutput .= "<input accept=\".pdf,.doc,.docx\" etc----/>\n";
This helps with Google Chrome, but Safari, Firefox ignore the modifications
Preventing users to submit the form if any other type of file is detected would be the ideal solution.
Thank you
I would recommend using a third party file uploading library like Plupload. We always use ValidForm Builder together with Plupload; works like a charm.
However you can use the meta array to implement custom attributes in the <input> tag without having to hardcode anything:
$objForm->addField(
"upload-document",
"Upload Document",
ValidForm::VFORM_FILE,
array(), // Validation array
array(), // Error handling array
array( // Meta array
"fielddata-extensions" => "pdf,doc,docx"
)
);
By prefixing meta keys with the 'field' prefix, you add that specific meta to the <input> field itself instead of it's wrapping <div class='vf__optional'></div>
The above example will output:
<div class="vf__optional">
<label for="upload-image">Upload Image</label>
<input type="hidden" name="MAX_FILE_SIZE" value="2097152">
<input type="file" value="" name="upload-image[]" id="upload-image" class="vf__file" data-extensions="pdf,doc,docx">
</div>
So using a combination of meta and a third party file upload handler, you can actually to pretty cool stuff.
That being said -- I must admit that the file upload field didn't get as much attention as the other field types lately.
I encounter a problem when testing form validation with angularjs
According to angularjs form guide,
an input control that has the ngModel directive holds an instance of NgModelController. Such a control instance can be published as a property of the form instance using the name attribute on the input control.
I created test code at plunker, it all works fine untill I change the input name from
<input type="number" name="age" ng-model="user.age" max="100" required>
<p>{{form1.age.$error}}</p>
to
<input type="number" name="user[age]" ng-model="user.age" max="100" required>
<p>{{form1.user[age].$error}}</p>
Does this mean angular can not recognize array syntax in form input?
The problem for me is I want to keep the normal form submission flow and only use angular for form validation, so I need to keep form input as array to work with the backend form handling
It has nothing to do with Angular. It is a syntactic JS error.
If you want to reference a property named user[age], you should do it like this:
form1['user[age]'].$error
form1.user[age] is incorrectly interpreted as (form1.user)[age]
I am somewhat of a novice with CGI Perl and am working on a web app that uses 'mode' and 'action' variables to determine which pages load.
$mode = param('mode');
$action = param('action');
if ($mode eq 'page1') {
if ($action 'eq') {
&performAction;
}
displayPage1;
}
elsif ($mode eq 'page2') {
&displayPage2
}
During development I have been having trouble figuring out the best way to set these variables when trying to navigate to different modes/actions after a form submit.
In some cases, putting a hidden value in the form will work
hidden(-name=>'action',-value=>'save')
but sometimes it will not. In case of the latter, putting param('action',"save") before the form will make the action change when the form is submitted.
I am unable to figure out why this happens though, are there factors that affect these two variables that I am unaware of?
What I now need to do is have two buttons on the same form, one which will just set the action to save the form data, and another which will save the form data but navigate to another mode/page with that form data.
If anyone could at least point me in the right direction for what I should be researching I would be greatly appreciative.
By default the CGI module implements a state-preserving behavior called "sticky" fields. The way this works is that if you are regenerating a form, the methods that generate the form field values will interrogate param() to see if similarly-named parameters are present in the query string. If they find a like-named parameter, they will use it to set their default values.
You want
hidden(-name=>'action', -value=>$new_value, -override=>1)
or
hidden(-name=>'action', -value=>'default_value')
param('hidden_name', $new_value);
This is a try , not sure if it would work.
Try setting hidden variable before button and changing it before every button, so the new value should be taken.
For ex:
<input type='hidden' name='op' value='save'/>
<input type='submit' name='Save Form' value='SaveForm'/>
<input type='hidden' name='op' value='submit'/>
<input type='submit' name='Submit Form' value='SubmitForm'/>
<input type='hidden' name='op' value='cancel'/>
<input type='submit' name='Cancel Form' value='CancelForm'/>
You can check for hidden variable 'op' in perl script.
This has driven me really bananas. It's so simple and easy and yet I can't figure out what's wrong with it.
I want to get my checkbox value populated in my controller (for testing purposes).
Here is my form.
<a href='#' name='submitForm'>submit the form</a>
//I have jquery attached to this tag and will submit the form when user clicks it
echo form_open('test/show');
echo form_checkbox('checkbox[]','value1');
echo form_checkbox('checkbox[]','value2');
echo form_checkbox('checkbox[]','value3');
echo form_checkbox('checkbox[]','value4');
echo "<input type='text' name='text1' value='ddd'>";
echo form_close();
//My controller test
public function show(){
$data1=$this->input->post('text1');
//I can get text1 value from input box
$data2=$this->input->post('checkbox');
//it keeps giving me undefined index 'checkbox'
$data3=$_POST['checkbox'];
//same error message
//WTH is going on here!!!!!
}
Please help. This thing drives me nuts! Thanks.
UPDATE:
Thanks for the help. To be more precisely, my submit button is a <a> tag and outside of form tag. It appear that I have to include <a> tag inside my form tag to make them works. Is that true?
A checkbox will not submit any data if it is unchecked as they're not considered successful (as per the w3c specification here)
If you actually tick the box and submit, it'll work - in fact it does, I've just tested it.
You need to wrap calls to $_POST in the isset() function.
if( isset( $_POST['checkbox'] ) ) {}
Calling $this->input->post('checkbox') shouldn't give you an undefined index error as the method deals with this eventuality. the Input::post() method returns false or the value of the checkbox.
Edit --
In response to your amendment to your question, you must use an element of type input with the type attribute set to submit in order to submit your form data without the use of Javascript etc. This button must be INSIDE the <form></form> which you are intending to submit.
<input type="submit" value="Submit">
The type="submit" causes the browser to send the data as submit event occurs. If you wish to use another element insider or outside of the form to do this you need to use Javascript. This however can be disabled on a per browser/user basis and isn't reliable as a result.
// Standard Javascript
<form name="myform"...
<a onclick="javascript:document.myform.submit();" href="javascript:void(0)">Submit</a>
// jQuery
$('#my-a-tag-submit-button').live( 'click', function() {
$('#my-form').submit();
}