JQuery’s mythical being strategies extremely created straightforward to post information a knowledge an information and come back that data while not refreshing the page. We are able to apply this jQuery mythical being post in CodeIgniter still. Before continuing any we tend to take into account that you just square measure a touch at home with CodeIgniter PHP framework. With the assistance of Associate in Nursing example you'll find out how to post information to a controller victimization the CodeIgniter jQuery mythical being methodology.
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Step-1. We create a html code in index.php
<!DOCTYPE html>
<html>
<head>
<title>Registration | SoftwareQuery</title>
</head>
<body>
<body>
<form id="form" method="post">
<table>
<tr>
<td>Name:</td>
<td><input type="text" name="name" id="name"/>
</tr>
<tr>
<td>Email:</td>
<td><input type="email" name="email" id="email">
</tr>
<tr>
<td colspan="2">
<input type="button" name"submit" id="submit">
</td>
</tr>
</table>
<span id="p1" style="display:none;"> Success</span>
<span id="p2" style="display:none;">Error</span>
</form>
</body>
</html>
-----------------------------------------------JQuery Code-------------------------------------------
Step- 2. Search html code ctrl+F this tag </head> and put the script.
<script src="https://ajax.googleapis.com/ajax/libs/dojo/1.13.0/dojo/dojo.js"></script>
-----------------------------------------<Script>...................................................................................
Step-3. Go got html code and press button ctrl+f search </body> tag put this script.
<script>$(function(){
$('#p1').hide();
$('#p2').hide();
$("#submit").click(function(){
var empty = false;
$('#name').each(function() {
if ($(this).val().length == 0) {
empty = true;
}
});
if (empty) {
$('#p2').fadeIn();
$("#p2").delay(2000).fadeOut();
}
else {
var empty = false;
$('#email').each(function() {
if ($(this).val().length == 0) {
empty = true;
}
});
if (empty) {
$('#p2').fadeIn();
$("#p2").delay(2000).fadeOut();
}
else{
var name = $("input[name=name]").val();
var email = $("input[name=email]").val();
$.ajax({
type: "POST",
url: "<?php echo base_url('controoler_name/insert/'); ?>",
data: {name:name,email:email},
success: function() {
$('#p1').fadeIn();
$("#p1").delay(2000).fadeOut();
$('#form').find("input[name=name], input[name=email]").val("");
}
})
}
})
</script>
----------------------------------------------Create Model---------------------------------------------
Step-4. Go go to Model Folder and open create new model file save any name the put this code.After put code go to browser and type localhost/phpmyadmin create new database name then make two field email,name.
function insert($name,$email)
{
$data = array(
'column_name'=> $name,
'coulmn_name_email' => $email
);
$this->db->insert('table_name',$data);
return true;
}
.............................................................Create Controller...................................................................
Step-5. Go go to controller Folder and open new model file save any name the put this code.
function insert(){
$name= $this->input->post('cname');
$email= $this->input->post('cemail');
if(($cname=='')||($cemail=='')){
echo 'error';
}else
$data = $this->model->insert($name,$email);
if($data == 'true'){
echo 'success';
else
{
echo 'error';
}
}
}
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