CODEWITHSUNDEEP
phpjqueryajaxyii
Antonio López Esclapez
Antonio López Esclapez

Reputation: 15

Send data through ajax to controller in yii

First of all thanks all for your help, it's very useful. Im starting with yii and Im a bit lost yet.

I have create a jquery script where I validate a form and then I send it to my controller to work with it and save in the db.

But Im doing it wrong I think I cant connect with my controller. Here is the code:

Jquery script(after all the validate stuff, the variables are fine):

$.ajax({
            type: "POST",
            url: "<?php echo Yii::app()->request->baseUrl; ?>/proceso/guardarproceso",
            data:
            {
            post_nombre: nombre,
            post_empresa: empresa,
            post_fechaI: fechaI,
            post_fechaF: fechaF,
            post_descripcion: descripcion
            },
            success: function(result)
            {
            alert(result);
            }
        });

And my controller ProcesoController:

public function actionGuardarProceso(){

            $nombre = $_POST['post_nombre'];
            $empresa = $_POST['post_empresa'];
            $fechaI = $_POST['post_fechaI'];
            $fechaF = $_POST['post_fechaF'];
            $descripcion = $_POST['post_descripcion'];

            echo $nombre;
    }

Im not working with the db yet, I only want to see if I have done it well and the alert(result) shows me the content of $nombre, but instead of that the alert shows me all the html code of the view(yes all xD)

I have done it too:

public function accessRules()
    {
        return array(
            array(
                  'allow',
                  'actions'=>array('index','guardarproceso'),
                  'users'=>array('*'),
            ),
       );
    }

But nothing...

Anyone cuold help me or give me some idea? Thank you all again

Upvotes: 1

Views: 7655

Answers (3)

Ionut Flavius Pogacian
Ionut Flavius Pogacian

Reputation: 4801

1st error: url: "<?php echo Yii::app()->request->baseUrl; ?>/proceso/guardarproceso",

replace guardarproceso with guardarProceso

2nd error:

$.ajax({
  type: "POST",
  url: url,
  data: data,
  success: success,
  dataType: dataType
});

dataType is missing; it should be json

read here http://api.jquery.com/jquery.post/

also, in the controller action actionGuardarProceso, use:

echo json_encode(array('key'=>$nombre)); exit

Upvotes: 1

Dhanush Bala
Dhanush Bala

Reputation: 1132

 $.ajax({
                type: "POST",
                url: "<?php echo Yii::app()->request->baseUrl; ?>/proceso/guardarproceso",
                data:
                {
                post_nombre: nombre,
                post_empresa: empresa,
                post_fechaI: fechaI,
                post_fechaF: fechaF,
                post_descripcion: descripcion
                },
            success:  function(result) {
               alert(result);
            },
            error:function(jqXHR, textStatus, errorThrown){
                alert('error::'+errorThrown);
            }
            });

first try this you will get post data or not 

public function actionGuardarProceso(){
echo "<pre>";
print_r($_POST);
exit;

    }

if not getting any data , try with your model

  public function actionGuardarProceso(){
        $model = new modelname;
        echo $_POST['modelname']['post_nombre'];
       exit;

       }

else try with the following 

 public function actionGuardarProceso(){
           echo Yii::app()->request->getPost("post_nombre");
         exit;

           }

i hope you will get now from the above any methods

Upvotes: 0

Chris
Chris

Reputation: 219

If a ajax or jquery result prints the html of the page its usually a error in the url. check your network jumps to see if it goes to the controller action you want. also baseurl is slower than createUrl. try Yii::app()->createUrl and edit that until it goes to the correct destination.. But the base of my theory is that your url is incorrect.

Upvotes: 0

Related Questions