checkbox data not insert in mysql using codeigniter - mysql

I try to insert my checkbox data in CodeIgniter. but data did not inserted in the database.
here is my view file:
<input type="checkbox" name="feature[]" value="WIFI" >
<input type="checkbox" name="feature[]" value="TV">
I am trying to use implode to convert the array into the string, but then I don't how to add in $data array, so they inserted in together
here is my controller:
public function save()
{
$this->load->model('Partner_model');
$feature = $this->input->post('feature');
$fea=array(
'feature'=>json_encode(implode(",",$feature))
);
$user_data= array(
'pname' => $this->input->post('pname'),
'type' => $this->input->post('type'),
'address' => $this->input->post('address'),
'about' => $this->input->post('about'),
'city' => $this->input->post('city'),
'code' => $this->input->post('code')
);
if($this->Partner_model->save($user_data,$fea))
{
$msg = "save sucesss" ;
}
else
{
$msg = "not save";
}
$this->session->set_flashdata('msg', $msg);
$this->load->view('partner_profile');
}
& here is my model:
public function save($data,$fea)
{
return $this->db->insert('property', $data,$fea);
}

Your model is faulty.
You are passing three arguments to insert() but the third you use is not appropriate.
That argument should be a boolean that indicates whether to escape values and identifiers or not. You need to incorporate $fea into $data which should probably be done in the controller.
There is an easier way to create the array $user_data since it is essentially a copy of $_POST just use $this->input->post().
Also, there is no obvious reason why you use json_encode. Unless you need it that way when you retrieve it from the DB there is no reason to bother with it. Consider removing json_encode.
First, change the model
public function save($data)
{
return $this->db->insert('property', $data);
}
Here's a revised save method
public function save()
{
$this->load->model('Partner_model');
$user_data = $this->input->post(); //makes a copy of $_POST
$feature = $this->input->post('feature');
if($feature) //because $feature will be null if no boxes are checked
{
$user_data['feature'] = json_encode(implode(",", $feature));
}
$msg = $this->Partner_model->save($user_data) ? "save sucesss" : "not save";
$this->session->set_flashdata('msg', $msg);
$this->load->view('partner_profile');
}
An explanation as requested via comments.
A call to $this->input->post('pname') returns the value of $_POST['pname'] if it is exists, but returns null if it does not exist.
When you create $user_data you make six calls to $this->input() with a different "key" each time to make a copy of $_POST.
$this->input->post() without any arguments returns the whole $_POST array. (See documentation)
$user_data = $this->input->post();
Makes a copy of $_POST using one line of code. It will include $_POST['feature'] if any boxes are checked, but $_POST['feature'] will not be set if no boxes are checked.
There are two ways to test if any boxes were checked. First we can test if isset($_POST['feature']) == true or we can test if $this->input->post('feature') == true. I use the second with the call
if($feature)
Which is pretty much the same as any of the following lines
if($feature != false)...
if($feature != null)...
if( ! empty($feature))...
if( ! is_null($feature))...
In other words, if($feature) evaluates as true if $feature is set and is anything except null, false, 0, "0", "" (an empty string), array() (an empty array)

public function save()
{
$this->load->model('Partner_model');
$feature = $this->input->post('feature');
$user_data= array(
'pname' => $this->input->post('pname'),
'type' => $this->input->post('type'),
'address' => $this->input->post('address'),
'about' => $this->input->post('about'),
'city' => $this->input->post('city'),
'code' => $this->input->post('code'),
'feature'=>json_encode(implode(",",$feature))
);
if($this->Partner_model->save($user_data)){
$msg = "save sucesss" ;
}else{
$msg = "not save";
}
$this->session->set_flashdata('msg', $msg);
$this->load->view('partner_profile');
}
model file should be :
public function save($data) {
return $this->db->insert('property', $data);
}

Related

How to retrive from other table and input it again into another table in laravel?

I want to retrieve some data from 'tbl_karyawan' and input into 'tbl_absen' which is if the NIP exist from 'tbl_karyawan' then parshing some data into 'tbl_absen'. I was create the code, and the data is going well. but i have something trouble with
i want the data input in Nip_kyn be like 'KIP001' not [{"Nip_kyn":"KIP001"}].
this is my Model
public function presensi($data)
{
$isExist = DB::table('tbl_karyawan')
->where('Nip_kyn', $data)->exists();
if ($isExist) {
$namakyn = DB::table('tbl_karyawan')->where($data)->get('Nama_kyn');
$nippppp = DB::table('tbl_karyawan')->where($data)->select('Nip_kyn')->get($data);
$values = array('Nip_kyn' => $nippppp, 'Nama_kyn' => $namakyn, 'Jam_msk' => now(), 'Log_date' => today());
DB::table('tbl_absen')->insert($values);
} else {
echo 'data not available';
}
}
this is my controller
public function get()
{
$day = [
'time_in' => $this->AbsenModel->timeIN(),
'time_out' => $this->AbsenModel->timeOut(),
'break' => $this->AbsenModel->break(),
// absen here
's' => $this->AbsenModel->absensi(),
];
$data = [
'Nip_kyn' => Request()->Nip_kyn,
];
$this->AbsenModel->presensi($data);
return view('v_absen', $data, $day);
}
Yup, I finally got it, the problem is on my model.
$nama_karyawan = DB::table('tbl_karyawan')->where($data)->value('Nama_kyn');
$nipkyn = DB::table('tbl_karyawan')->where($data)->value('Nip_kyn');
I just change 'get' to 'value'.

How to force data update on laravel validation custom unique

I'm new in laravel. I have a table with menu_id and title I tried to make this title field unique when have the same menu_id. I found the solution here
But I got problem when update it. Can anyone help please?
My code
Validator::extend('unique_custom', function ($attribute, $value, $parameters)
{
// Get the parameters passed to the rule
list($table, $field, $field2, $field2Value) = $parameters;
// Check the table and return true only if there are no entries matching
// both the first field name and the user input value as well as
// the second field name and the second field value
return \DB::table($table)->where($field, $value)->where($field2, $field2Value)->count() == 0;
});
public function updateSubmenu( Request $request) {
$this->validate( $request, [
'menu_id' => 'required',
'title' => 'required|unique_custom:posts,title,menu_id,'.$request->menu_id,
'order_by' => 'required|integer',
'description' => 'required'
],
[
'title.unique_custom' => 'This title already token'
]
);
}
Can you explain what problem have you got on update? Some exception?
Edit:
If you couldn't update record if title not changes, you need to add one condition to Validator:
Validator::extend('unique_custom', function ($attribute, $value, $parameters)
{
// Get the parameters passed to the rule
list($table, $field, $field2, $field2Value) = $parameters;
// If old value not changed, don't check its unique.
$current = \DB::table($table)->where('title')->first();
if( $current->{$field} == $value) {
return true;
}
return \DB::table($table)->where($field, $value)->where($field2, $field2Value)->count() == 0;
});

API Json in symfony form

i'm creating an application in symfony 2.8 (with php5.4) and in my form (that i'm building) i want to display a list of a projects through an API in json format.
Now i'm stuck, and i don't know how to do this.
I know the database of the API there is a table "projects" and i want target the column 'name' to display names of projects
here's my code:
/**
* #Route("/form")
*/
public function formAction(Request $request)
{
$url = 'https://website.com/projects.json';
$get_json = file_get_contents($url);
$json = json_decode($get_json);
$form = $this->createFormBuilder()
->add('Project', 'choice') // <-- ???
->add('send', 'submit' ,array('label' => 'Envoyer'))
->getForm();
$form->handleRequest($request);
return $this->render('StatBundle:Default:form.html.twig', array('form' => $form->createView(), 'project' => $json));
}
You can pass the choice as second argument as array as example:
$jsonAsArray = json_decode($get_json, true); // with true return an array
$builder->add('Project', 'choice', array(
'choices' => $jsonAsArray,
// *this line is important, depends of the data*
'choices_as_values' => false,
));
More info in the doc here.
Hope this help

How to avoid the instruction "return" inside a function in Symfony2?

I would like to know how to avoid the instruction "return" inside a function in Symfony2. In other words how can I make a void function which doesn't return anything. In fact I have tried that for a long time but every time I run the code I did I see this error message: "The controller must return a response" ... By the way, this is the code that I have:
public function AddeventsgroupeAction(Request $request) {
$eventg = new eventsgroupe();
$form = $this->createForm(new eventsgroupeType(), $eventg);
$em = $this->getDoctrine()->getManager();
$securityContext = $this->get('security.context');
$token = $securityContext->getToken();
$user = $token->getUser();
$id = $user->getId();
$groupe=$this->getRequest('groupe');
$idg = intval($groupe->attributes->get('id'));
$qb = $em->createQueryBuilder();
$qb->select('l')
->from('IkprojGroupeBundle:Groupe', 'l')
->from('IkprojGroupeBundle:eventsgroupe', 'e')
->where(' l.id = :g and e.idGroupe = l.idAdmin and l.id = e.idEventGroupe');
$qb->setParameter("g", $idg);
$query = $qb->getQuery();
$res = $query->getResult();
$rows = array();
foreach ($res as $obj) {
$rows[] = array(
'id' => $obj->getId());
}
if ($request->isMethod('POST')) {
$form->handleRequest($request);
if ($form->isValid()) {
$eventg-> setIdGroupe($id);
$eventg-> setIdEventGroupe($idg);
$em->persist($eventg);
$em->flush();
return $moslem="yes";
}
} else {
return $this->render('IkprojGroupeBundle:GroupeEvents:Addeventgroupe.html.twig', array(
'groupe' => $rows,
'event' => $eventg,
'form' => $form->createView(),
));
}
}
How can I replace the instruction : return $moslem="yes"; in order to not return anything??...Is that possible??
To answer your basic question, a simple return will return a void from your function.
The "controller must return a response" message actually comes from the request handler. You need to tell the request handler what you want it to do. There is no default page so a void return will trigger the error.
In most cases, after successfully processing a posted form you will want to return a redirect response.
Something like:
$form->handleRequest($request);
if ($form->isValid()) {
...
$em->flush();
return $this->redirect($this->generateUrl('task_success'));
I should point out that your form code seems to be from S2.1 or older. It's unnecessarily complicated. You should be using at least 2.3. Make sure you are looking at the correct version of the documentation. Hint: the isValid() takes care of the POST check.
http://symfony.com/doc/current/book/forms.html#handling-form-submissions
It's also worth while to understand the request/response workflow.
http://symfony.com/doc/current/book/http_fundamentals.html
Digging into the code can also help in understanding where the error message is coming from:
Symfony\Component\HttpKernel\HttpKernel#handleRaw($request)
Simple, delete the else statement and if $request->isMethod('POST') or $form->isValid() returns false the code inside will not be executed then the script return the default view.
EDIT: you can also make a redirect with a flash message where needed like this:
$this->get('session')->getFlashBag()->add('success', 'your success message');
return $this->redirect($this->generateUrl('your_route'));
Remember to add support for flash message in your view looking at the Symfony2 docs

insert if not exists Codeigniter

my controller:
function getFeed()
{
$feed_url = $this->input->get("url");
$content = file_get_contents($feed_url);
$x = new SimpleXmlElement($content);
foreach($x->channel->item as $entry) {
$feeds[] = array(
'title' => (string)$entry->title,
'url'=> (string)$entry->link,
'username' => $this->session->userdata('username')
);
$this->load->model('membership_model');
$this->membership_model->feeds($feeds);
}
Model:
function feeds($feeds_data)
{
$this->db->insert_batch('feeds', $feeds_data);
}
Is there a function to insert if only the row doesn't exists in the table? I have a table with 4 column : id,title,url,username. I have an anchor when i click him it calls geFeed function and insert the info into table. But i want to insert only if not exists.
I had the same challenge, so i eventually come up with a function which might be helpful to you.
function safe_update_batch($table_name,$records,$filter_field)
{
$filters=array();
foreach($records as $record)$filters[]=$record[$filter_field];
$this->db->query("SET SESSION group_concat_max_len=10000000");
$query=$this->db->select("GROUP_CONCAT($filter_field) AS existing_keys",FALSE)->where_in($filter_field, $filters)->get($table_name);
$row=$query->row();
$found_fields=explode(',',$row->existing_keys);
$insert_batch=array();
$update_batch=array();
foreach($records as $record)
{
if(in_array($record[$filter_field],$found_fields))$update_batch[]=$record;
else $insert_batch[]=$record;
}
if(!empty($insert_batch))$this->db->insert_batch($table_name,$insert_batch);
if(!empty($update_batch))$this->db->update_batch($table_name,$update_batch,$filter_field);
}
//sample usage
$this->safe_update_batch('feeds', $feeds_data,'title');
You can try this in your model!!
function insertClient($array)
{
$this->db->from('MyTable');
$this->db->where('Id', $array['Id']);
$query = $this->db->get();
if($query->num_rows() != 0){
$data = array(
'name'=>$array['name'],
'phone'=>$array['phone'],
'email'=>$array['email']
);
$this->db->where('Id', $array['Id']);
$this->db->update('CLIENTS', $data);
}else{
$data = array(
'name'=>$array['name'],
'phone'=>$array['phone'],
'email'=>$array['email']
);
$this->db->insert('CLIENTS',$data);
}
}
In controller:
$this->site_model->insertClient($_POST);
Sadly if you are using the active record class an INSERT IF NOT EXISTS function doesn't exist. You could try
Extending the active record class (easier said than done)
You could set indexes on certain columns as UNIQUE so that MySQL will check to see if it already exists
You could do some kind of SELECT before your INSERT to determine if the record is already there
For the queries where you need to do INSERT IF NOT EXISTS do $this->db->query('INSERT IF NOT EXISTS...')
function getFeed()
{
// Load the model up here - otherwise you are loading it multiple times
$this->load->model('membership_model');
$feed_url = $this->input->get("url");
$content = file_get_contents($feed_url);
$x = new SimpleXmlElement($content);
foreach($x->channel->item as $entry) {
// check if the feed is unique, if true then add to array
if( $this->membership_model->singleFeedIsUnique($entry) == TRUE ){
$feeds[] = array(
'title' => (string)$entry->title,
'url'=> (string)$entry->link,
'username' => $this->session->userdata('username')); }
} //foreach
// check to make sure we got any feeds with isset()
// if yes, then add them
if (isset($feeds)){ $this->membership_model->feeds($feeds); }
}
You can try this in your model and leave you controller without changes
function feeds($feeds_data)
{
$data = array(
title => $feeds_data[0],
url => $feeds_data[1],
username => $feeds_data[2]
);
$this->db->select('*');
$this->db->from('mytable');
$this->db->where('title',$feeds_data[0]);//you can use another field
if ($this->db->count_all_results() == 0) {
$query = $this->db->insert('mytable', $data);//insert data
} else {
$query = $this->db->update('mytable', $data, array('title'=>$feeds_data[0]));//update with the condition where title exist
}
}
you can check the id if you have it, adding in the data array and use it to check if exist