Getting the value of a row in CodeIgniter - php

I am trying to get the value of a row in the database but not working out that well. I not sure if can work like this.
I am just trying to get the value but also make sure it from the group config and the key.
Model
function getTitle() {
return $this->db->get('setting', array(
->where('group' => 'config'),
->where('key' => 'config_meta_title'),
->where('value'=> ) // Want to be able to return any thing in the value
))->row();
}
In the controller I would do this:
function index() {
$data['title'] = $this->document->getTitle();
$this->load->view('sample', $data);
}

First, you have this line set to this:
$data['title'] $this->document->getTitle();
That should be an = assignment for $this->document->getTitle(); like this:
$data['title'] = $this->document->getTitle();
But then in your function you should actually return the setting value from your query with row()->setting:
function getTitle() {
return $this->db->get('setting', array(
->where('group' => 'config'),
->where('key' => 'config_meta_title'),
->where('value'=> ) // Should return this row information but can not.
))->row()->setting;
}
But that said, I am unclear about this:
->where('value'=> ) // Should return this row information but can not.
A WHERE condition is not a SELECT. It is a condition connected to a SELECT that allows you to SELECT certain values WHERE a criteria is met. So that should be set to something, but not really sure what since your code doesn’t provide much details.

Found Solution Working Now. For Getting Single Item From Database In Codeigniter
Loading In Library
function getTitle($value) {
$this->CI->db->select($value);
$this->CI->db->where("group","config");
$this->CI->db->where("key","config_meta_title");
$query = $this->CI->db->get('setting');
return $query->row()->$value;
}
Or Loading In Model
function getTitle($value) {
$this->db->select($value);
$this->db->where("group","config");
$this->db->where("key","config_meta_title");
$query = $this->db->get('setting');
return $query->row()->$value;
}

Related

Passing data between 2 models in 1 controller function laravel

I need to get the lecture id in my student table and from then only can i get the lecture's data using that id. The problem is, the $lectureID is null when i try to apply it on the lecture model, but when i check on the console, it does get the data. Thanks in advance.
Controller
public function getLecture($id)
{
$lectureID = student::select('lecture_id_FK')->where('student_id',$id)->first();
$lectureDATA = lecture::where('lecture_id',$lectureID)->first();
return $lectureDATA;
}
you can try two ways
public function getLecture($id)
{
## way 1
$lectureID = student::where('student_id',$id)->value('lecture_id_FK');
$lectureDATA = lecture::where('lecture_id',$lectureID)->first();
## way 2
$lectureID = student::where('student_id',$id)->first();
$lectureDATA = lecture::where('lecture_id',$lectureID->lecture_id_FK)->first();
return $lectureDATA;
}
before check what inside $lectureID:
public function getLecture($id)
{
$lectureID = student::select('lecture_id_FK')->where('student_id',$id)->first();
dd($lectureID);//I believe inside will be object with lecture_id_FK
//then just
$lectureDATA = lecture::where('lecture_id',$lectureID->lecture_id_FK)->first();
return $lectureDATA;
}
Recommended way to do it:
In Student model class ( recommend use capital S )
class student
{
public function lecture(){
return $this->hasOne('App\lecture','lecture_id_FK','lecture_id');
}
}
then just load studen with lecture like this
$student = studend::with('lecture')->find($id);
$lecture = $student->lecture;

Call to a member function insert() on null. Codeigniter

Hi everyone I am getting the following error when I submit my form for my CI 3 website:
Fatal error: Call to a member function insert() on null
This error is occurring on line 20 which is:
$query = $this->db->insert('temp_subscribed_users', $data);
Here is the full function:
public function add_temp_user($key)
{
echo "hello";
$data = array(
'TEMP_EMAIL' => $this->input->post('email'),
'TEMP_KEY' => $key
);
echo var_dump($data);
$query = $this->db->insert('temp_subscribed_users', $data);
if($query)
{
return true;
}else{
return false;
}
}
I am not sure what it means by null. The table name is correct and I did a var_dump to confirm that the array is being populated. I also made sure that I am getting into the function by echoing "hello" and it is outputting onto the page.
Any help is appreciated thank you!
Additional info: Running using XAMPP localhost.
load database and then call insert function. Codeigniter does not load database automatically for the performance issue.
$this->load->database();
$query = $this->db->insert('temp_subscribed_users', $data);
Well first of all you're not utilizing the MVC model of codeigniter. Controller is for functions, Model is for the database connections.
First autoload your database, If not just put it in the code. But here is how it should look like.
CONTROLLER FUNCTION
public function add_temp_user($key)
{
echo "hello";
$this->load->model('MY_MODEL');
//If you're not autoloading db include the next line
//$this->load->library('database');
$data = array(
'TEMP_EMAIL' => $this->input->post('email'),
'TEMP_KEY' => $key
);
echo var_dump($data);
//If you confirmed the data var dumped
$success = $this->MY_MODEL->insert_to_db($data);
if($success == true)
{
//Do something
}
else
{
//Do something
}
}
MODEL
public function insert_to_db($data)
{
$query = $this->db->insert('temp_subscribed_users', $data);
//
if($query)
{
return true;
}
else
{
return false;
}
}
Make sure the TEMP_EMAIL and TEMP_KEY are the columns in your database and temp_subscribed_users is your table name
Try to solve your problem by getting into the autoload.php in the config folder and add database on the array for libraries, like this: $autoload['libraries'] = array('database');
Please check whether object is created or not.
Check that object is available in that class

Loop to get user data

I can't put real code here because is very long and will be hard to
explain.
I have users table in database and I have data table in database too.
So, to get the user data I'll pass user_id as parameter. Like this:
public function get_user_data($user_id) {
}
But. I can only get 1 data per "request". (Keep reading)
public function user_data() {
$getUsers = $this->db->get('users');
foreach($getUsers->result_array() as $user)
{
$data = $this->get_user_data($user->ID);
var_dump($data); // Only return 1 data;
}
}
But, I guess that have an way to "bypass" this but I don't know. I'm having trouble thinking.
As I said, I want to "bypass" this, and be able to send multiple user IDs, my real function do not accept that by default and can't be changed.
Thanks in advance!
replace
foreach($getUsers->result_array() as $user)
{
$data = $this->get_user_data($user->ID);
var_dump($data); // Only return 1 data;
}
to this
foreach($getUsers->result_array() as $user)
{
$data[] = $this->get_user_data($user->ID);
}
var_dump($data);
If you are aiming at sending more data to the function, you always need to make signature change of your function as one of the below :
function get_user_data() {
$args = func_get_args();
/** now you can access these as $args[0], $args[1] **/
}
Or
function get_user_data(...$user_ids) {
/** now you can access these as $user_ids[0], $user_ids[1] **/
}
// Only higher version of PHP
But I am not sure how you will handle returning data.
EDIT: Yes, then in the function, you can collect data in array and return an array of data from function.
If you can change in your function from where to where_in I think you will get an easy solution.
public function get_user_data($user_ids)
{
// your db code
$this->db->where_in('ID',$user_ids); //replace where with where_in
}
public function user_data()
{
$getUsers = $this->db->get('users');
foreach($getUsers->result_array() as $user)
{
$user_ids[] = $user->ID;
}
$this->get_user_data($user_ids);
}

How to access a certain GET data in CakePHP?

I am currently writing an adress book and using a framework (CakePHP) an MVC for the first time. Unfortunately I have some trouble.
I want to realize the following:
In case the URL is
/contacts/view/
I want to show all contacts in a list. In case there is an id given after /view/, e.g.
/contacts/view/1
I just want to display the contact with the id 1. (complete different view/design than in the first case)
My ContactsController.php is the following
public function view($id = null){
if(!$this->id){
/*
* Show all users
*/
$this->set('mode', 'all');
$this->set('contacts', $this->Contact->find('all'));
} else {
/*
* Show a specific user
*/
$this->set('mode','single');
if(!$this->Contact->findByid($id)){
throw new NotFoundException(__('User not found'));
} else {
$this->set('contact', $this->Contact->findByid($id));
};
}
}
But "$this->mode" is always set as "all". How can I check whether the id is set or not?
I really want to avoid "ugly" URL-schemes like ?id=1
Thanks in advance!
Your code is only meeting the if part and its not going to else part. Use (!$id)..
$_GET data is retrieved through the URL. In CakePHP this means it's accessed through that method's parameters.
I'm arbitrarily picking names, so please follow! If you're in the guests controller and posting to the register method you'd access it like this
function register($param1, $param2, $param3){
}
Each of these params is the GET data, so the URL would look something like
www.example.com/guests/param1/param2/param3
So now for your question How can I check whether the id is set or not?
There are a couple of possibilities. If you want to check if the ID exists, you can do something like
$this->Model->set = $param1
if (!$this->Model->exists()) {
throw new NotFoundException(__('Invalid user'));
}
else{
//conduct search
}
Or you can just search based on whether or not the parameter is set
if(isset($param1)){ //param1 is set
$search = $this->Model->find('all','conditions=>array('id' => $param1)));
}
else{
$search = $this->Model->find('all');
}
You should only change the conditions not the whole block of code like
public function view($id = null){
$conditions = array();
$mode = 'all';
if($id){
$conditions['Contact.id'] = $id;
$mode = 'single';
}
$contacts = $this->Contact->find('all', array('conditions' => $conditions));
$this->set(compact('contacts', 'mode'));
}

Form Validation w/ sql + codeigniter

I'm working on creating a callback function in codeigniter to see if a certain record exists in the database, and if it does it'd like it to return a failure.
In the controller the relevent code is:
function firstname_check($str)
{
if($this->home_model->find_username($str)) return false;
true;
}
Then in the model I check the database using the find_username() function.
function find_username($str)
{
if($this->db->get_where('MasterDB', array('firstname' => $str)))
{
return TRUE;
}
return FALSE;
}
I've used the firstname_check function in testing and it works. I did something like
function firstname_check($str)
{
if($str == 'test') return false;
true;
}
And in that case it worked. Not really sure why my model function isn't doing what it should. And guidance would be appreciated.
if($this->home_model->find_username($str)) return false;
true;
Given that code snippet above, you are not returning it true. If that is your code and not a typo it should be:
if($this->home_model->find_username($str)) return false;
return true;
That should fix it, giving that you did not have a typo.
EDIT:
You could also just do this since the function returns true/false there is no need for the if statement:
function firstname_check($str)
{
return $this->home_model->find_username($str);
}
So the solution involved taking the query statement out of if statement, placing it into a var then counting the rows and if the rows was > 0, invalidate.
Although this is a more convoluted than I'd like.
I find your naming kind of confusing. Your model function is called 'find_username' but it searches for a first name. Your table name is called 'MasterDB'. This sounds more like a database name. Shouldn't it be called 'users' or something similar? I'd write it like this :
Model function :
function user_exists_with_firstname($firstname)
{
$sql = 'select count(*) as user_count
from users
where firstname=?';
$result = $this->db->query($sql, array($firstname))->result();
return ((int) $result->user_count) > 0;
}
Validation callback function :
function firstname_check($firstname)
{
return !$this->user_model->user_exists_with_firstname($firstname);
}

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