I'm a rookie on codeigniter and I'm trying to solve a problem where I try to get some images from the database using a foreach cicle
I have this on my controller to get the images
$data['products'] = '.base_url('$this->Cart_model->get_img()').';
and this on my model
public function get_img()
{
$sql = "SELECT image FROM products";
$query = $this->db->query($sql);
return $query->result_array();
}
In my view it get me a syntax error beacause something is wrong in the controller
Already checked the View.tpl and everything is fine, must be something in the controller
You forgot the concatenation operator to concatenate the strings.
$data['products'] = '.base_url(' . $this->Cart_model->get_img() . ').';
Related
I am trying to filter a database by having a category column, so I search for the category and the correct items are returned. However some items have more than one category so I have used comma separated values to put two or more categories in one column. I tried this method:
public function get_cat_article($value)
{
$table = $this->get_table();
$this->db->where('cat', $value);
return $this->db->get($table);
}
but that only picks up the first of the separated values. I tried using 'where_in' but that also picked up only the first value. I tried 'like' and that worked but was not definitive. A search for 'WS' included other categories like 'IWS'. I googled for a solution and fin_in_set was suggested so I tried:
public function get_cat_article($value)
{
$table = $this->get_table();
$this->db->where('find_in_set('.$value.',cat)');
return $this->db->get($table);
}
But this didn't work at all. Have I got the syntax wrong or is this the wrong approach?
Have you tried like query? In a quick demo, this works for me.
public function get_cat_article($value)
{
$table = $this->get_table();
$this->db->like('cat', $value); // say, $value = 'reebok'
return $this->db->get($table);
// Produces: SELECT * FROM $table WHERE `cat` LIKE '%reebok%' ESCAPE '!'
}
For more details read CodeIgniter official user guide here.
Try like method but this way:
public function get_cat_article($value)
{
$table = $this->get_table();
$this->db->like('cat', $value . ',', 'after');
$this->db->or_like('cat', ',' . $value . ',');
$this->db->or_like('cat', ',' . $value, 'before');
return $this->db->get($table);
}
The problem appears to be solved using the following line of code using the MySql 'find_in_set' technique. I got the code from this website at Check the conditions in comma separated values
and the key line is:
$this->db->where("FIND_IN_SET( '$post_period' , period) ");
I just got the syntax wrong.
I use this code to display data according to the user who is logged in, and display their data.
but with this syntax the data that appears does not match the defined data, and also other user data sometimes appears too.
$id=$this->session->userdata('ses_id');
$this->db->SELECT('*');
$this->db->from('doc_priv_std');
$this->db->join('acc',' doc_priv_std.id_acc=acc.id_acc');
$this->db->where('doc_priv_std.id_acc ',$id,('and
type_doc LIKE "%.doc%"
or "%.docx%"
or "%.pdf"
or "%.xls%"
or "%.xlsx%"
or "%.ppt%"
or "%.pptx%"
or "%.zip%"
or "%.rar%"'));
$query = $this->db->get();
return $query->result_array();
please help so that the data that appears as defined.
Your syntax is wrong - you can't do this with your way - you have to organise your Query properly - try the following
$id=$this->session->userdata('ses_id');
$arrExtension = ['.docx', '.pdf', '.xls', '.xlsx', '.ppt', '.pptx', '.zip', '.rar'];
$this->db
->select('*')
->from('doc_priv_std')
->join('acc',' doc_priv_std.id_acc=acc.id_acc')
->where('acc.id_acc', $id)
->group_start();
foreach($arrExtension AS $strExtension)
{
$this->db->or_like('type_doc', $strExtension);
}
$query = $this->db->group_end()->get();
return $query->result_array();
Use var_dump($this->db->last_query()); before your return to see the db query you are actually running to help with debugging, but it could be that you where statement is wrong? Try this instead:
$where = 'WHERE type_doc LIKE "%.doc%"
or "%.docx%"
or "%.pdf"
or "%.xls%"
or "%.xlsx%"
or "%.ppt%"
or "%.pptx%"
or "%.zip%"
or "%.rar%"';
$this->db->select('*')
->join('acc',' doc_priv_std.id_acc=acc.id_acc')
->where('doc_priv_std.id_acc', $id)
->where($where);
$query = $this->db->get();
return $query->num_rows() > 0 ? $query->result_array() : FALSE;
I am trying to join two tables and return an array in my Model method in CodeIgniter with php. I've gone through a few previously posted similar questions on stackoverflow and modified my code accordingly. But they don't seem to work. Hence would love to know what's wrong with the following.
I'm using the following method but am currently getting exceptions. Would appreciate suggestions in this regard.
Model Method
public function getUserDetails($username)
{
$uid = $this->getUserUid($username);
$this->db->select('s.uid, s.name,s.phone, s.studentId, s.type, sp.profiledesc, sp.age');
$this->db->from('sysuser as s');
$this->db->join('studentprofile as sp', 's.uid = sp.uid', 'left');
$this->db->where(array('s.uid' => $uid));
$query = $this->db->get();
return $query->result();
}
Controller
$data1['details'] = $this->userModel->getUserDetails($username);
$this->load->view('studentDashboard/viewProfile',$data1);
View
...
<h2>
<?php foreach($details as $detail){?>
<?php echo $detail->s.name;?>
<?php }?>
</h2>
...
In the view, I've also tried just echoing $detail->name but this doesn't work either.
At first, use print_r($details) for checking your data. If it's returning anything or not.
Then echo your value like this $detail['name']
Fixed Code:
public function getUserDetails($username)
{
$uid = $this->getUserUid($username);
$this->db->select("*");
$this->db->from('sysuser');
$this->db->join('studentprofile', 'studentprofile.uid = sysuser.uid');
$this->db->where('sysuser.uid',$uid);
$query = $this->db->get();
return $query->result();
}
Look, i´m not sure that i understood your code, what means this line
$uid = $this->getUserUid($username);
You´re calling a method and sending the name to retrieve the userid, right?
I´ll write that method like you should have it:
public function getUserid($user){
$this->where->id($user);
return $this->get('whatever table')->row();
//i think you forgot this ->row()
}
then
public function getUserDetails($username)
{
$uid = $this->getUserUid($username);
//here already you should bring with ->row() already
//you can use this var_dump here to confirm too
//var_dump($uid);
//exit;
$this->db->select('s.uid, s.name,s.phone, s.studentId, s.type, sp.profiledesc, sp.age');
//line´s right
//the from method is disposable, so i put it into the get but here it´s right too
$this->db->join('studentprofile as sp', 's.uid = sp.uid', 'left'); //ok
$this->db->where($uid); //this line is the wronger, i´ve made the fix
//as this is an where, you´d bring only one value with ->row() from that other method
$query = $this->db->get('sysuser as s');
//the 'from' i putted here, just to write a line less
return $query->result();
when you need to test what you´re returng do a var_dump here
//commenting the return above
//$test = $query->result();
//var_dump($test);
}
I'm trying to create a function that runs a query that returns all of the data located in my MySQL database.
My current code only returns the one row of data (there are 7)
function staff_get() {
$this->load->database();
$sql = 'SELECT * from Staff';
$query = $this->db->query($sql);
$data = $query->row();
$this->response($data, 200);
}
I'd imagine it has something to do with the line "$data = $query->row();" however I've tried switching "row" with "array" but this doesn't work. The text is designed to come out as plaintext so that I can manipulate it using a jQuery template.
Thank you for your help in advance.
You need to encase the results in a while loop. Something along the lines of this.
function staff_get() {
$this->load->database();
$sql = 'SELECT * from Staff';
$query = $this->db->query($sql);
while($data = $query->row()) {
$this->response($data, 200);
}
}
I'm trying to submit an update function, but for some reason it's not working and I can't figure out why... Someone who does?
UPDATE SQL-SYNTAX:
public function updateProject($db, $id) {
$sql = "UPDATE tblProject SET
name = '".$db->escape($this->name)."',
photo1 = '".$db->escape($this->photo1)."'
WHERE id = '".$id."'";
return $db->insert($sql);
}
INSERT FUNCTION:
public function insert($sql) {
mysql_query($sql, $this->_connection);
return mysql_affected_rows($this->_connection);
}
PHP:
$project = new Project();
$project->name = $_POST['newproject_name'];
$project->photo1 = $_FILES['images']['name'][0];
if($project->updateProject($_DB, $projectname)) {
$feedback = "OK!";
} else {
$feedback = "NOT OK!!";
}
And in case you were wondering, $project->name and $project->photo1 are filled in correctly.
Any ideas? I hope I gave you everything you need, if not, let me know!
EDIT 1: I used the 2 first answers, but no results. Yet...
EDIT 2: I also don't get anything from $feedback
It looks like you have a stray opening parenthesis after the SET keyword. Remove it.
public function updateProject($db, $id) {
$sql = "UPDATE tblProject SET
name = '".$db->escape($this->name)."',
photo1 = '".$db->escape($this->photo1)."'
WHERE id = '".$id."'";
return $db->insert($sql);
}
public function updateProject($db, $id)
requires 2 parameters being passed but when you do
if($project->updateProject($_DB))
you're only passing 1??
Your updateProject needs two variables and the second one is missing in your call, resulting in an invalid query.
Edit: Based on your edit; I'm guessing $id needs to be an integer or a string, you are passing an object.
Exactly what row do you want to update? I don't see anything in your php code that defines the ID of the row you want to modify, you are just generating a new object, not getting one from a database for example.