In my model function contain variables from controller.But that value got from controller.But it is not get into model.
model
function get_sub_marks_data($division,$subj_name)
{
$sql = "SELECT student_name,".$subj_name." AS marks FROM f_tbl WHERE
division='".$division."' ORDER BY student_name asc";
echo $sql;
$query=$this->db->query($sql);
return $query;
}
controller
Post data from ajax
function get_subject_wise_marks()
{
$subj_name=$this->input->post('sub');
$exam=$this->input->post('exam');
// $classid=6;
$division='A';
$subj_name = strtolower($subj_name);
if($exam == 't1'||$exam == 't2')
{
$subj_name= $exam.'_10_'.$subj_name;
}
else if($exam == 't3'|| $exam == 't4')
{
$subj_name= $exam.'_20_'.$subj_name;
}
$sub_marks=$this->IM->get_sub_marks_data($division,$subj_name);
}
$subj_name the value is not get into model..
Any mistakes in this..
Have you considered to load the model before
$sub_marks=$this->IM->get_sub_marks_data($division,$subj_name); ???
Cause I don't see any loading of the model in your code. So try this :
...
$this->load->model('IM');
$sub_marks=$this->IM->get_sub_marks_data($division,$subj_name);
Tell me if that worked for you
Have you loaded model IM in your constructor
If not you need to load it like
$this->load->model('IM');
before calling its function(before below line)
$sub_marks=$this->IM->get_sub_marks_data($division,$subj_name);
Related
I want to display a value in view in CodeIgniter but I am getting several errors like trying to get the property on non-object. I think my code is correct but I am getting errors. Below is my code.
controller:
public function trainer($id)
{
$user_record = $this->db->query("select * from usr_data where usr_id=$id")->result();
$data['title'] = 'trainer Dashboard';
$data['user_record'] = null;
$data['active_courses'] = [];
$data['inprogress'] = [];
if(count($user_record)) {
$user_record = $user_record[0];
$data['title'] = ucwords($user_record->firstname).' Dashboard';
$data['user_record'] = $user_record;
$active_courses = $this->base_model->getTrainercourseAll($id);
$data['active_courses'] = $active_courses;
$inprogress = $this->base_model->getstaffinprogress($id);
$data['inprogress'] = $inprogress;
}
$this->load->view('trainer-dashboard', $data);
}
model:
public function getstaffinprogress($user_id) {
$result=$this->executeSelectQuery("select AVG(m.percentage) from object_data o, ut_lp_marks m where o.obj_id=m.obj_id and o.type='crs' and m.status=1 ");
return $result;
}
view:
<h3>Avg inprogress:<?php echo "<span style='color:#ff00ff;font-family:verdana;'>".$inprogress->percentage."</span>";?></h3>
I want to display the column percentage which is coming from database.above code is in the controller, model and view.i thought my controller code is wrong.
Anyone help me to get rid of this error. I want to display a value in view in CodeIgniter but I am getting several errors like trying to get the property on non-object. I think my code is correct but I am getting errors. Below is my code.
Try this in your view file,
if(isset($inprogress)){
echo $inprogress->percentage;
}
Then your code look like this,
<h3>Avg inprogress:<?php if(isset($inprogress)){ echo "<span style='color:#ff00ff;font-family:verdana;'>".$inprogress->percentage."</span>";}?></h3>
Then call the controller function. I think inprogress is not set at the first time.
If it doesn't work, try to var_dump($inprogress) in controller and check value and type.
And try this code in your model. Query also seems not correct
public function getstaffinprogress($user_id) {
$this->db->select_avg('ut_lp_marks.percentage');
$this->db->where('ut_lp_marks.obj_id', $user_id);
$this->db->where('object_data.type', 'crs');
$this->db->where('ut_lp_marks.status', 1);
$this->db->join('object_data', 'object_data.obj_id = ut_lp_marks.obj_id');
$query = $this->db->get('ut_lp_marks');
return $query->result_array();
}
I assume that your db is ut_lp_marks. Then var_dump array and check data is correct first. Then access array element.
public function getstaffinprogress($user_id) {
$result = array();
$query=$this->db->query("select AVG(m.percentage) from object_data o, ut_lp_marks m where o.obj_id=m.obj_id and o.type='crs' and m.status=1 ");
foreach($query->result() as $row){
$result = $row;
}
return $result;
}
Also check $inprogress->percentage exists before print in view.
Undefined variable: data in my view
This is a simple display data in the input.
So, why this input isn't display my query result at it?
my view
<input type="text" name="sitename" value="<?php echo $data['sitename']; ?>"><br>
model
public function getData()
{
$query = "SELECT * FROM $this->tablename ORDER BY 'id' DESC LIMIT 1";
if (!$sqli = mysqli_query($this->cxn->connect(),$query))
{
throw new Exception("Error Processing Request");
}
else
{
$num = mysqli_num_rows($sqli);
while ($num > 0)
{
$data = mysqli_fetch_array($sqli);
$num--;
}
return $data;
}
}
Simply because a variable is declared somewhere, doesn't mean it is available everywhere. All variables have scope in which they are accessible. See this: http://php.net/manual/en/language.variables.scope.php for more information on scope.
You need to pass the $data variable into your view. I image you're using some sort of MVC framework since you have a model and a view. If this is the case you can lookup how to pass variables into views in that specific framework. The basic structure of your controller method might look something like this:
//sudo code - not specific to an actual framework
public function controller_method()
{
$data = $model->getData();
$this->template->set('data',$data);
$this->template->load('view');
}
Just search how to do that in your specific framework. Hope that helps!
EDIT
Base on your comment it looks like you're setting data after you load the view. You need to swap the order and call $display = new Display("main"); $data = $display->getData(); before you include'../model/display.php';
If the query returns 0 rows, your while() loop will never execute, so it won't set $data.
Since you're only returning 1 row from the query, you don't need a loop, you can just use an if. Then you can return $data only when it succeeds.
public function getData()
{
$query = "SELECT * FROM $this->tablename ORDER BY 'id' DESC LIMIT 1";
if (!$sqli = mysqli_query($this->cxn->connect(),$query))
{
throw new Exception("Error Processing Request");
}
else
{
if ($data = mysqli_fetch_array($sqli))
{
return $data;
}
else
{
return null;
}
}
}
Hi I have problem when i tried to save attribute of model to database. I write in OctoberCMS and i have this function:
public function findActualNewsletter()
{
$actualNewsletter = Newsletter::where('status_id', '=', NewsletterStatus::getSentNowStatus())->first();
if (!$actualNewsletter) {
$actualNewsletter = Newsletter::where('send_at', '<=', date('Y-m-d'))->where('status_id', NewsletterStatus::getUnsentStatus())->first();
$actualNewsletter->status_id = NewsletterStatus::getSentNowStatus();
dd($actualNewsletter);
}
return $actualNewsletter;
}
getSentNowStatus()=2;
getUnsentStatus()=1;
dd($actualNewsletter) in my if statement show that status_id = 2 But in database i still have 1. I used this function in afterSave() so i dont need:
$actualNewsletter->status_id = NewsletterStatus::getSentNowStatus();
$actualNewsletter->save();
becosue i have error then i use save in save.
Of course i filled table $fillable =['status_id']. And now i dont know why its not save in database when it go to my if. Maybe someone see my mistake?
If you are trying to modify the model based on some custom logic and then save it, the best place to put it is in the beforeSave() method of the model. To access the current model being saved, just use $this. Below is an example of the beforeSave() method being used to modify the attributes of a model before it gets saved to the database:
public function beforeSave() {
$user = BackendAuth::getUser();
$this->backend_user_id = $user->id;
// Handle archiving
if ($this->is_archived && !$this->archived_at) {
$this->archived_at = Carbon\Carbon::now()->toDateTimeString();
}
// Handle publishing
if ($this->is_published && !$this->published_at) {
$this->published_at = Carbon\Carbon::now()->toDateTimeString();
}
// Handle unarchiving
if ($this->archived_at && !$this->is_archived) {
$this->archived_at = null;
}
// Handle unpublishing, only allowed when no responses have been recorded against the form
if ($this->published_at && !$this->is_published) {
if (is_null($this->responses) || $this->responses->isEmpty()) {
$this->published_at = null;
}
}
}
You don't have to run $this->save() or anything like that. Simply modifying the model's attributes in the beforeSave() method will accomplish what you desire.
I have a view (myView) in which there is a form. The form action is myController/myFunction1 which is used to validate the input variables in the form and insert it to the database by calling a model function. This works perfectly fine.
Now, I need a dropdown box inside the form, for which the values will be fetched from a table (called business) in the db.
This is the code I wrote in my model to fetch the values
public function get_dropdown_list() {
$this -> db -> select('business_name');
$result = $this -> db -> get('business');
if ($result -> num_rows() > 0) {
foreach ($result->result_array() as $row) {
$new_row['value'] = htmlentities(stripslashes($row['business_name']));
$row_set[] = $new_row;
}
}
return $row_set;
}
I'm not entirely sure if this is correct.
What I need to know is, if this is correct, what should be the code inside the controller and the view to display the result as a dropdown in the form in the myView.
And if this model itself is wrong, how do I get it working?
P.S. : I'm new to CodeIgniter. I have been going through S.O and various other sites to get this thing working for quite a bit of time now. This might seem to be a repeated question for which I'm really sorry, because I could not find a solution from the already available discussions dealing with the same issue. Any help is very much appreciated.
try Model :-
public function get_dropdown_list() {
$this -> db -> select('business_name');
$result = $this -> db -> get('business');
if ($result -> num_rows() > 0) {
return $result->result_array();
}
else {
return false;
}
}
Controller :-
1. include model in your controller
2. call the function and send data to view.
$this->load->model('model_name');
$this->data['dropdown'] = $this->model_name->get_dropdown_list();
$this->load->view('yourview', $this->data);
get value in view:-
print_r($dropdown)
Loop your data and make a dropdown
<select name="dropdown">
<?php foreach($dropdown as $d) {?>
<option value="<?php echo $d;?>"><?php echo $d;?></option>
<?php }?>
</select>
Call This function in controller for getting your records from DB
$data['records'] = $this->my_model->get_data();
In my_model.php
function get_data()
{
$query = "select * from my_tab";
$res = $this->db->query($query);
if ( $res->num_rows )
{
return $res->row_array();
}
return false;
}
In view.php
<select>
<?for($i=0;$i<count($records);$i++)
{
?>
<option>$records[$i]->name</option>
<?php } ?>
</select>
I am creating a web-service backend for a mobile app I am developing. (I am an experience Obj-C developer, not a web-designer!) Essentially I would like to use Codeigniter and Phil Sturgeon's RESTful API Server https://github.com/philsturgeon/codeigniter-restserver however, I'm having some trouble getting it all setup and working.
I have MySQL database that is setup with data in it. And I need some help writing a CodeIgniter PHP model and controller that returns JSON data of what is inside that database. All of the tutorials and forum post's i have found deal with hardcoded data in the controler, not a MySQL database. I would ideally like to have the URL formatted like this http://api.mysite.com/v1/search?id=1&name=foo&city=bar , I could potentially have 50+ parameters to pass in the url.
Using Phil's code, I have come up with this as my controller:
public function index_get()
{
if (!$this->get('id'))
{
$this->response(NULL, 400);
}
$data = $this->grid_m->get_id($this->get('id'));
if ($data)
{
$this->response($data, 200);
}
else
{
$this->response(NULL, 404);
}
}
That only gets me one search term: id?=# .. I need to know how to get multiple search terms
Here is my Codeigniter model:
<?php
class Grid_m extends CI_Model
{
function get_all()
{
$query = $this->db->get('grid');
if ($query->num_rows() > 0)
{
return $query->result();
}
return FALSE;;
}
This just returns EVERYTHING in my MySQL database regardless of what id or url term I pass it in the URL.
I'm a big noob when it comes to developing my own custom API so any suggestions on how to fix my controller and database model would be a huge help!
Thanks for the help!
-brian
This is old question but if somebody go here and still need help, try these code:
In your controller:
public function index_get()
{
$where = '';
if ($this->get('id')) // search by id when id passed by
{
$where .= 'id = '.$this->get('id');
}
if ($this->get('name')) // add search by name when name passed by
{
$where .= (empty($where)? '' : ' or ')."name like '%".$this->get('name')."%'";
}
if ($this->get('city')) // add search by city when city passed by
{
$where .= (empty($where)? '' : ' or ')."city like '%".$this->get('city')."%'";
}
// you can add as many as search terms
if ( ! empty($where))
{
$data = $this->grid_m->get_searched($where);
if ($data)
{
$this->response($data, 200);
}
else
{
$this->response(NULL, 404);
}
}
else
{
$this->response(NULL, 404);
}
}
In your model, create get_searched function:
class Grid_m extends CI_Model
{
function get_searched($where = NULL)
{
if ( ! is_null($where))
{
$this->db->where($where);
$query = $this->db->get('grid');
if ($query->num_rows() > 0)
{
return $query->result();
}
}
return FALSE;
}
}
User Codeigniter's Active record to build proper query, you can build any type of query using methods of active record, refer following example I have just added one condition in it you can add more conditions as per your need.
<?php
class Grid_m extends CI_Model
{
function get_all()
{
$this->db->select('col1, col2, col3')->where('id', 5);
$query = $this->db->get('grid');
if ($query->num_rows() > 0)
{
return $query->result();
}
return FALSE;;
}
}
Please check you query
$query = $this->db->get_where('grid', array('id' => $id));