如何用AJAX发送单选按钮值并用PHP恢复它


How to send radio button value with AJAX and recover it with PHP

我一直试图发送用户选择的单选按钮的值,并用php恢复该值,但问题是我无法恢复该值。这是我的代码:

HTML

<form name="submission" action="">
    <input type="radio" name="ex1" id="ex1_a" value="1">
    <input type="radio" name="ex1" id="ex1_b" value="2">
    <input type="radio" name="ex1" id="ex1_c" value="3">
    <button class="buttonS" type="submit"> 
        Submit
    </button>
</form>

带AJAX的查询脚本

$(function() {   
           $(".buttonS").click(function() {  
           // validate and process form here
           var radio_button_value;
           if ($("input[name='ex1']:checked").length > 0){
               radio_button_value = $('input:radio[name=ex1]:checked').val();
           }
           else{
               alert("No button selected, try again!");
               return false;
           }
           $.ajax({
               type: "POST", 
               url: "save.php",  
               data: radio_button_value,  
               success: function() { 
                    alert("form submitted: "+ radio_button_value);
               }
            });
            return false;
         });
});

PHP

<?php
         $selected_button = $_POST['ex1'];
         echo "Test";
         echo $selected_button;
?>

AJAX部分似乎在显示警报后工作,但我不知道它是否正确发送了值,或者php是否错误,显示了echo"Test",但echo$selected_button从未出现。如果有任何帮助,我将不胜感激。

在ajax函数中,您必须指定一个参数名称和值:

$.ajax({
               type: "POST", 
               url: "save.php",  
               data: {"ex1":radio_button_value},  
               success: function() { 
                    alert("form submitted: "+ radio_button_value);
               }
            });