Codeigniter中的AJAX找不到基本URL

Question

And my other question is that, is the code correct, will my AJAX request get to my controller ?

Here is the relevant parts on the code:

My view (I have this on a POST form cause I want to send the data on another table) :

<script> var base_url = <?php echo base_url(); ?> </script>

<label for="exampleInputEmail1">Apartament</label>
<select onchange="showUser(this.value)" name ="txtApartament1" class="form-control">
    <?php foreach($getEntry as $value) { ?>
    <option><?php echo $value->apartament ?></option>
    <?php }?>
</select>

In the same view this is my AJAX part:

function showUser(str) {
    if (str == "") {
        document.getElementById("txtApartament1").innerHTML = "";
        return;
    } else { 
        if (window.XMLHttpRequest) {
            // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("txtApartament1").innerHTML = this.responseText;
            }
        };
        xmlhttp.open("GET", base_url + "usercontroller/ajaxp?q="+str,true);
        xmlhttp.send();
    }
}

And this is my controller - usercontroller.php function:

public function ajaxp(){
echo "received";
}

I will get this error:

(index):257 Uncaught ReferenceError: base_url is not defined at showUser (localhost/adminigniter1/:257:29) at HTMLSelectElement.onchange (localhost/adminigniter1/:205:89)

My project filename is codeigniter1!

I did not set any config file on codeigniter. should I ? Anything else?

Answer 1

Try like this..

1.Load url helper using $this->load->helper('url') or in applicaiton/config/autoload.php.

2.Set base_url config item in application/config/config.php

$config['base_url'] = 'your_url';

3.In your script.Remove var .Because it creates variable as local .So you can not use inside function.

base_url = <?php echo base_url();

OR put it var base_url = <?php echo base_url(); ?> var base_url = <?php echo base_url(); ?> inside showUser() function.

UPDATE

var url = base_url + "usercontroller/ajaxp?q="+str; //OR  var url = <?php base_url();?>+"usercontroller/ajaxp?q="+str;

console.log(url);

xmlhttp.open("GET",url,true);
xmlhttp.send();