Get the radio button value through ajax to php file

问题

After clicking the radio button, the value from the radio button is not being passed when the onclick event is triggered. Here is my code:

<form name="Form1" id="color" style="font-size: 100%" action="#">  
    <input type="radio"  name="radio1"  id="radio1" onclick = "MyAlert()" value="blue"/>Blue  <br /></p>  
<p> <input type="radio"  name="radio1" id="radio1" onclick = "MyAlert()" value="red"/>Red  
</form>  

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">  
function MyAlert()  {  
    var radio1=$('input[type="radio"]:checked').val();
    var pass_data = {
            'radio1' : radio1,
        };
        alert(pass_data);
        $.ajax({
            url : "",
            type : "POST",
            data : pass_data,
            success : function(data) {
            }
        });
        return false;
    }  
</script>
<?php
    echo $radio1=$_GET['radio1'];
?>

When I click the radio button, I get the error

Undefined index: radio1

I want to display value of the radio button when clicking it within the same page.

回答1:

Firstly make ajax to separate PHP page where you will access the radio value. Also make alert after you receive the data.

$.ajax({
    url : "post.php",
    type : "POST",
    data: pass_data,
    success : function(data) {
        // alert radio value here
        alert(data);
    }
});

Crete a separate PHP file post.php where you access radio input. Since you are making POST request you need to use $_POST instead of $_GET to get radio button value.

<?php 
    $radio1 = $_POST['radio1'];
    echo $radio1;
?>

回答2:

  <input type="radio"  id="status" name="status" value="1" /> Mbyllur<br />
  <input type="radio"  id="status" name="status" value="0" /> Hapur<br />


  function MyAlert()  
      {  
  var radio1=$('input[type="radio"]:checked').val();
  var pass_data = {
                    'radio1' : $('input[name=status]:checked').val(),
              };
              alert(pass_data);
              $.ajax({
                    url : "",
                    type : "POST",
                    data : pass_data,
                    success : function(data) {
                    }
              });
              return false;
      } 

回答3:

1. I would use a newer version of jquery .

2. You can't give two elements the same id.

3. I would rewrite the code as follow :

$(function() {
	$(document).on('change', '[name="radio1"]' , function(){
  	var val = $('[name="radio1"]:checked').val();
    alert(val);

    
    /* 

    Ajax code 1 (GET) : 
    $.get('/myurl?val='  + val, function(){

    });


    Ajax code 2 (POST) : 
    $.post('/myurl', {val : val},  function(){

    });

    */
  }); 

});

<form name="Form1" id="color" style="font-size: 100%" action="#" >  
<input type="radio"  name="radio1" value="blue"/>Blue  <br />
<p> <input type="radio"  name="radio1" value="red"/>Red  
</form>  
<script src="https://code.jquery.com/jquery-1.12.3.min.js"></script>

回答4:

Try This -->

<form name="Form1" id="color" style="font-size: 100%" action="#" >  
    <input type="radio"  name="radio1"  id="radio1" onclick = "MyAlert()" value="blue"/>Blue  <br /></p>  
    <p> <input type="radio"  name="radio1" id="radio1" onclick = "MyAlert()" value="red"/>Red  
    </form>



<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">  
function MyAlert()  
    {  
var radio1=$('input[type="radio"]:checked').val();
//alert(radio1);
var pass_data = {
            'radio1' : radio1,
        };
        //alert(pass_data);
        $.ajax({
            url : "request.php",  // create a new php page to handle ajax request
            type : "POST",
            data : pass_data,
            success : function(data) {
            }
        });
        return false;
    }  
</script>

request.php

<?php
if(isset($_POST['radio1']))
{
    echo $radio1=$_POST['radio1'];
}
?>

Above code handle with ajax so, its not refresh the page.

回答5:

<script>
    $(document).ready(function() {
        $("#Enviar").click(function (e) {

            var cedula = document.getElementById("Cedula").value;
            var Nombre = document.getElementById("Nombre").value;
            var Apellido = document.getElementById("Apellido").value;
            var Sexo = $('input:radio[name=SexoC]:checked').val();
            var Edad = document.getElementById("Edad").value;
            var FechaN = document.getElementById("date").value;
            var Tele = document.getElementById("tel").value;
            var Direccion = document.getElementById("Direccion").value;
            var Invitacion = document.getElementById("Invitacion").value;
            var CasaG = document.getElementById("CasaG").value;
            var Rango = document.getElementById("Rango").value;

            var cadena = "Cedula="+cedula+"&Nombre="+Nombre+"&Apellido="+Apellido+"&Sexo="+Sexo+"&Edad="+Edad+"&Fecha="+FechaN+"&Tele="+Tele+"&Direccion="+Direccion+"&Invitacion="+Invitacion+"&CasaG="+CasaG+"&Rango="+Rango;

                    $.ajax({
                        type:'POST',
                        url:'datos/Registrar.php',
                        data: cadena,
                        beforeSend: function(){
                        console.log(cadena);
                     },
                        success:function(Resp){
                            alert(Resp);
                        }
                    });
                    return false;

        });
    });
        </script>

来源：https://stackoverflow.com/questions/37314341/get-the-radio-button-value-through-ajax-to-php-file