store form data in mysql database using ajax in codeigniter - php

Question

Edit:
It was a problem with localhost and I think with the htaccess file. Although I couldn't make it in localhost, the script is running fine on the web host.
I want to store my form data in to the database using ajax in codeigniter. The problem is that everything is fine, except I'm getting a 500 server internal error.
My contoller:
public function order()
{
$order = $this->main_model->order($_POST);
if($order)
{
return true;
}
else
{
return false;
}
}
my model:
function order($options = array())
{
$options = array(
'client_Name' => $this->input->post('oName'),
'client_Phone' => $this->input->post('oPhone')
);
$this->db->insert('md_orders', $options);
return $this->db->insert_id();
}
and of course I'm using stepsForm script and this is the js code I have:
var theForm = document.getElementById( 'theForm' );
new stepsForm( theForm, {
onSubmit : function( form ) {
var form_data = {
oName: $('#oName').val(),
oPhone: $('#oPhone').val(),
};
$.ajax({
url: "<?php echo base_url() . 'main/order/'; ?>",
type: 'POST',
data: form_data,
success: function(msg) {
alert(msg);
}
});
}
} );
and this is the HTML code:
<form id="theForm" class="simform" autocomplete="off">
<ol class="questions" id="questions">
<li>
<span><label for="oName">Your name:</label></span>
<input class="finput" id="oName" name="oName" type="text"/>
</li>
<li>
<span><label for="oPhone">Your Phone Number:</label></span>
<input class="finput ltr" data-validate="number" id="oPhone" name="oPhone" type="text"/>
</li>
</ol>
</form>
and the error I'm getting is :
POST http://localhost/123/main/order/ 500 (Internal Server Error)
Failed to load resource: the server responded with a status of 500 (Internal Server Error)
and nothing is stored in the database. What am I doing wrong?!

Answer 1

Try to check your input in controller before send it to model, although it is not mandatory. In your model you need just feed of your array. So you can just set name of parameter.
Your model something like this:
public function order($options) {
$data = array();
$data['client_Name'] = $options['oName'];
$data['client_Phone'] = $options['oPhone'];
$this->db->insert('md_orders', $data);
if($this->db->affected_rows() > 0) {
return $this->db->insert_id();
else {
return false;
}
}
Your controller something like this:
public function order() {
$order = $this->main_model->order($this->input->post());
if( $order !== false ) {
echo "Inserted!";
} else {
echo "Not inserted!";//These are your ajax success function msg parameter
}
}
Also, I can't see the logic of form parameter inside onSubmit property's function( form ). Maybe you could remove it?

Answer 2

In your view,place the following code in your script,please...
one_field = $('#one_field').val();
second_field = $('#second_field').val();
$.ajax({
type:"post",
data:{one_field:one_field, second_field:second_field},
url :'<?=base_url("directory/controller_name/your_method_name")?>',
success: function(data)
{
alert("success");
}
});
Now the place the following in your controller.
function your_method_name()
{
$your_data = array(
"table_field_1" => $this->input->post('one_field'),
"table_field_2" => $this->input->post('second_field')
);
$last_id = $this->your_model->insert_data_function($your_data);
if(isset($last_id)
{
//your code
}
else
{
//your code
}
}
In your Model,place the following.
public function insert_data_function($your_data)
{
$this->db->insert("your_table",$your_data);
return $this->db->insert_id(); //will return last id
}

Answer 3

Instead of using base_url() in your JS, you should try and use site_url()
Your CodeIgniter website runs through the index.php file.
The base_url() function will return the URL to your base directory.
The site_url() will return the URL to your index.php file.
Your form is trying to access "http://localhost/123/main/order" when it should be trying "http://localhost/123/index.php/main/order"
Hope this helps.
Edit
You should also have a look at the form helper.
https://ellislab.com/codeIgniter/user-guide/helpers/form_helper.html