how to get a passed data from the url in codeigniter -


hi have data passed in url http://www.test.com/dashboard/john-doe

i want page when enter http://www.test.com/dashboard/john-doe john doe name database retrieve when users can login.

so dashboard controller how able page if passed data john-doe next dashboard controller? can me figured thing out? muchly appreicated controller below

<?php if ( ! defined('basepath')) exit('no direct script access allowed');  class dashboard extends ci_controller {   public function __construct(){     parent::__construct();         $this->load->model('users_model', 'um');          $sessiondata = $this->session->userdata('loggedin');         $this->data['id'] = $sessiondata['userid'];          if(!$this->session->userdata('loggedin')){       redirect('login');     }    }      public function index($slug){         echo $slug; exit;         $this->data['title'] = 'dashboard | test';         $this->data['menu'] = 'dashboard';         //get customerinfo         $this->data['getcustomerinfo'] = $this->um->getcustomerinfo($this->data['id']);           $this->template_lib->set_view('templatelogin', 'dashboard', $this->data,'',$this->data);     } }

the index function loaded default if second segment of uri empty. if have parameters need define route:

$route['dashboard/(:any)'] = "dashboard/index/$1";

Comments

Post a Comment

Popular posts from this blog

php - failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request -

this code i'm using find latitude , longitude of place: function coords($address){ echo $url="http://maps.googleapis.com/maps/api/geocode/json?address=".$address; $json = file_get_contents(urlencode($url)); $data = json_decode($json, true); print_r($data['results'][0]['geometry']['location']['lat']); print_r($data['results'][0]['geometry']['location']['lng']); } however returns warning: failed open stream: http request failed! http/1.0 400 bad request if use above code in simple procedure works fine not in function. note: have tried curl_init()... method produced same result? instead of encoding whole url encode address part. following code work fine $add='jamshoro phase 2'; $url="http://maps.googleapis.com/maps/api/geocode/json?address=".urlencode($add); $json = file_get_contents($url); $data = json_decode($json, true); print_r($data['r
Read more

java - How to filter a backspace keyboard input -

i have code gives message when key number pressed: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochararray().length - 1]; if (!(ch >= '0' && ch <= '9')) { system.out.println("the char entered not number"); t.consume(); } } }); now if press on wrong button , press backspace remove it, error message. how can add backspace input if statement? i have found answer: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochar
Read more

java - Show Soft Keyboard when EditText Appears -

i have code in android app, alert dialog , keyboard, , when dialog shown want keyboard appear. not , not sure why? here code: final alertdialog.builder alert = new alertdialog.builder(this); alert.settitle("title"); alert.setmessage("message"); final edittext input = new edittext(this); input.setinputtype(inputtype.type_class_number); alert.setview(input); input.setonfocuschangelistener(new view.onfocuschangelistener() { @override public void onfocuschange(view v, boolean hasfocus) { if (hasfocus) { log.i(tag, "in focus"); inputmethodmanager inputmethodmanager = (inputmethodmanager) getsystemservice(context.input_method_service); inputmethodmanager.showsoftinput(input, inputmethodmanager.show_implicit); } } }); but keyboard not appear though edit text in focus, , tell appear why, doesn't come up? thanks in advance.
Read more