我希望能够在不提交表单的情况下将单选按钮的值发布到数据库,因此为什么我尝试使用“更改时”进行此操作。
$("input:radio[name=q1_MC]").on("change",function () {
var dunno1 = $(this).serialize();
$.ajax({
url: "get_response.php",type: "POST",data: dunno1,success: function (data) {
console.log("working)";
},error: function (request,status,error) {
console.log(request.responseText);
}
});
});
当我单击单选按钮之一时,console.log确实显示。
在get_response.php里面,我有:
<?php
require("db_connection.php");
if((isset($_POST['your_name']) {
$yourName = $conn->real_escape_string($_POST['your_name']);
$q1_MC = $conn->real_escape_string($_POST['q1_MC']);
$q2 = $conn->real_escape_string($_POST['q2']);
$q3 = $conn->real_escape_string($_POST['q3']);
$q4 = $conn->real_escape_string($_POST['q4']);
$q5 = $conn->real_escape_string($_POST['q5']);
$q6 = $conn->real_escape_string($_POST['q6']);
$q7_MC = $conn->real_escape_string($_POST['q7_MC']);
$q8 = $conn->real_escape_string($_POST['q8']);
$sql="INSERT INTO commenttable (name,q1_MC,q2,q3,q4,q5,q6,q7_MC,q8) VALUES ('".$yourName."','".$q1_MC."','".$q2."','".$q3."','".$q4."','".$q5."','".$q6."','".$q7_MC."','".$q8."')";
if(!$result = $conn->query($sql)){
die('There was an error running the query [' . $conn->error . ']');
} else {
echo "Thank you! Your feedback is appreciated";
}
}
?>
HTML:
<input type="radio" name="q1_MC" value="Excited"> Excited
<input type="radio" name="q1_MC" value="Optimistic"> Optimistic
<input type="radio" name="q1_MC" value="Indifferent"> Indifferent
<input type="radio" name="q1_MC" value="Nervous"> Nervous
<input type="radio" name="q1_MC" value="Sceptical"> Sceptical