Jquery和php不能在两个页面之间传递数据

Question

I have two pages in php and I need that the value of radioButton selection pass to another page, but it not work. The jquery code of the page that send the data is:

   $(document).ready(function(){ 
  $("#continue").click(function(){
var val = $("input[@name='opt']:checked").val();
        $.ajax({
            type: 'GET',
        url:'editor.php',
            data:'radio=' + val,
            dataTyoe:'html',
            succes: alert(val)
        });
    }); 
});

The html code into the php page is:

    <input type="radio" id="opt" name="opt" value="opt1" checked="checked">Opt 1<br/>
    <input type="radio" id="opt" name="opt" value="opt2"/>opt2<br />
    <input type="radio" id="opt" name="opt" value="opt3"/>opt3<br />

    <a href="editor.php" id="continue">Guardar continuar</a><br/>

And the code of the page that recive data is the follow.

<?php
$valor = $_REQUEST['radio'];
echo $valor

?>

Thanks

Answer 1

There were some spelling mistakes and need some correction. Try the following code;

$(document).ready(function(){

$("#continue").click(function(){
var val1 = $("input[name='opt']:checked").val();
    $.ajax({
        type: 'GET',
        url:'editor.php',
        data:'radio=' + val1,
        success: function(){
            alert(val1);
        }
    });
}); 

});

Answer 2

你拼错了成功：它应该是成功的：

Answer 3

You may have some errors : try replacing dataTyoe:'html' for dataType:'html' , Also $("input[@name='opt']:checked") by $("input[@name=opt]:checked") ... and the success callback too...

It would look like :

var val = $("input[@name=opt]:checked").val();
 $.ajax({
      type: "GET",
      url: "editor.php",
      data:'radio='+val,
      dataType: "html",
      async:false,
      success: function(){
         alert(val);
      }
   }
);

Also you could remove the link from the html :

  <a href="#" id="continue">Guardar continuar</a><br/>