表单需要在codeigniter中将数据发布到控制器

Question

I have a form,

<form>
<input id="amount"  type="text" placeholder="Give amount"  name="myamount"/>
<button class="btn btn-lg" type="submit" id="btn1" >FirstBTN</button> 
<button class="btn btn-lg" type="submit" id="btn2">SecondBTN</button>
</form>

Which needs to post data to a function inside the controller. I am trying to make two buttons, post data to two different controller functions. And this is my jquery code written in html file (view):

<script type="text/javascript">
$('#btn1').click(function() { 
        $.ajax({
            url: 'helloworld/firstmoney',
            type: 'POST',
            data: {'submit' : true ,myamount: $("#amount").val()},
            success: function (result) {

            }
        });  
 });

</script>

<script type="text/javascript">
    $('#btn2').click(function() { 
            $.ajax({
                url: 'helloworld/secondmoney',
                type: 'POST',
               data: {'submit' : true ,myamount: $("#amount").val()},
                success: function (result) {

                }
            });  

    });

    </script>

Here is the controller but I am not getting any values in my controller..

public function firstmoney(){
        $this->load->helper('form');

        $Money = $this->input->post("myamount");
        $data =  array(
        'Money' => $Money
         );
        $this->load->view("page1", $data);

    }

public function secondmoney(){
            $this->load->helper('form');

            $Money = $this->input->post("myamount");
            $data =  array(
            'Money' => $Money
             );
            $this->load->view("page2", $data);

        }

my controller name is helloworld and I need to assign values in $money but I cant get it working. I am a beginner programmer so please do tell me what steps I should follow.

Answer 1

First, your AJAX should look like this one.

    $('#btn2').click(function() { 
        $.ajax({
            url: 'helloworld/secondmoney',
            method: 'POST',//not type
            data: { myamount: $("#amount").val() }//no need for submit variable (although you can use it) since this will be submitted on click; no quotes over variables
        })
          .done(function(result)) {
              //do your stuff on response
          });
    });

Second: no need for ajax and you should use classic form with post method because you are making redirection in controller/method where data is posted to. AJAX requires JSON formated response, and it is not in this case. So wether change view to have form with method and action, wether change controller/method to return JSON string back to AJAX.

Answer 2

 var base_url= '<?php echo base_url() ?>';


 url: base_url+'helloworld /firstmoney',

Answer 3

You can try another way to do that

Try below code

form

<form action="" method="post" id="myform">
<input id="amount"  type="text" placeholder="Give amount"  name="myamount"/>
<button class="btn btn-lg" type="submit" id="btn1" >FirstBTN</button> 
<button class="btn btn-lg" type="submit" id="btn2">SecondBTN</button>
</form>

Then you can add jquery to change the form action on submit button clicks like below

<script type="text/javascript">

$('#btn1').click(function() { 
        $('#myform').attr('action', 'Your domain/Your controller/firstmoney'); 
 });

$('#btn2').click(function() { 
        $('#myform').attr('action', 'Your domain/Your controller/secondmoney'); 
 });

</script>

Now your form will call firstmoney() if you click #btn1 and it will call secondmoney() if you call #btn2

And you can get form data using

$this->input->post('myamount');

Answer 4

If you only want to print $_POST data then apply this code for btn1.

$(document).ready(function(){
    $('#btn1').click(function() { 
    event.preventDefault();

        $.ajax({ 

            url: "<?php echo base_url(); ?>" + "helloworld/firstmoney",         
            type: 'POST',
            data: {'submit' : true ,myamount: $("#amount").val()},
            success: function (result) { 
                $("body").html(result);
            }
        });  
    });
});

In your helloworld controller

public function firstmoney(){

    $Money = $this->input->post("myamount");
    $data =  array(
    'Money' => $Money
     );
    $data = $this->load->view("page1", $data, TRUE); //it will return that page and save it in $data 
}

In you page1.php view write the following code

<?php
echo $Money;
?>

Answer 5

<script type="text/javascript">
$('#btn1').click(function() { 
        $.ajax({
            url: 'helloworld/firstmoney',
            type: 'POST',
            data: {'submit' : true ,myamount: $("#myamount").val()},
            success: function (result) {

            }
        });  
 });

Keep the id and name value as myamount in above input tag and change in the jquery code also