CODEWITHSUNDEEP
phpmysqlcodeigniter
Darjeeling
Darjeeling

Reputation: 999

How can I get data from a MySQL database with Code Igniter?

This is my first time trying web programming and I'm currently trying to use Code Igniter. I got this from tutorial but my problem seems a little bit different, I tried several thing but none work

So I want to get data from my db, here's my code

public function can_log_in($username, $password){

    $query = $this->db->query("SELECT col1, col2 FROM table1 where id_login = '$username' and id_password = '$password'");

    $data = mysql_fetch_row($query); //got error this line, and several line also on libraries and helper
    if ($query->num_rows() == 1){
        return  (string) $data[0];
    }else {
        return "0";
    }
}

Error said:

Supplied argument is not a valid MySQL result resource

On controller I used:

if($this->model_get->can_log_in($username, $password) != "0")

Honestly I don't know if this work or should I use <> ? haven't got opportunity to tried it, but that's another problem.

So how can I get col1 and col2 here?

Upvotes: 1

Views: 730

Answers (4)

Kevin
Kevin

Reputation: 41885

Since you're using codeigniter, just be faithful to it, use its built-in functions:

public function can_log_in($username, $password)
{
    $username = $this->db->escape($username); // this automatically adds single quotes
    $password = $this->db->escape($password);
    $query = $this->db->query("SELECT col1, col2 FROM table1 where id_login = $username and id_password = $password");
    if($quer->num_rows() > 0) {
        $data = $query->row(); // fetches single row
        return $data->col1;
    }
}

I also suggest use codeigniters active record, so that you don't have to worry about escaping. They already to it for you:

public function can_log_in($username, $password)
{
    $this->db->select('col1', 'col2');
    $this->db->where('username', $username);
    $this->db->where('password', $password);
    $query = $this->db->get('table1');

    return ($query->num_rows() > 0);
}

In controller:

$username = $this->input->post('username');
$password = $this->input->post('password');
$can_log = $this->model_name->can_log_in($username, $password);
if($can_log) {
    // yahoo! can log!
} else {
    // sorry but you cannot!
}

Upvotes: 1

ANZAWI
ANZAWI

Reputation: 89

CodeIgniter provides several methods to perform on query results.

See here: https://ellislab.com/codeigniter/user-guide/database/results.html

result() returns an array of PHP objects.

row() returns a single PHP object for that row.

result_array() returns an array of arrays.

row_array() returns a single array for that row.

row() and row_array() optionally take a parameter which is the number of the row you'd like to return.

$rows = $query->result(); //array of objects
$rows_array = $query->result_array(); //array of arrays
$row = $query->row(); //single object
$row_array = $query->row_array(); //single array

Upvotes: 0

James
James

Reputation: 923

In PHP, you need to connect to a database using

$link = mysqli_connect("myhost","myusername","mypassword","mybd") or die("Error " . mysqli_error($link));

Once you've connected to a database you can query that database using 

$data = mysqli_query($con,"SELECT * FROM tablename");

Looking at what you've written, it seems like your code is hitting an error because you haven't connected to a database to query. Once you've established a connection you can submit queries and parse them using a loop to enumerate each row of results of 

for($i = 0;$i<mysqli_num_rows($data);$i++) {
    $row = mysqli_fetch_row($data));
    $username = row[0]; //usernames are column 1
    $password = row[1]; //passwords are column 2
    }

And so on. You can read up more on using PHP at PHP.net

Upvotes: 0

deviloper
deviloper

Reputation: 7240

Try this:

$query = $this->db->query("SELECT col1, col2 FROM table1 where id_login = '$username' and id_password = '$password'");

$data = $query->row_array();

if ($query->num_rows() == 1){
    return  (string) $data['col1']; // Need to give the field name
}else {
    return "0";
}

Upvotes: 0

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