Reputation: 35
I have to call this php file
<?php
session_start();
require 'connect.php';
//Prendo le tre variabili dalla form
if(isset($_POST['Titolo'])) {
$title_rew = $conn->real_escape_string($_POST['Titolo']);
}
if(isset($_POST['Voto'])) {
$voto_rew = $conn->real_escape_string($_POST['Voto']);
}
if(isset($_POST['Review'])) {
$review_rew = $conn->real_escape_string($_POST['Review']);
}
if(isset($_POST['ID_locale'])) {
$id_rew = $conn->real_escape_string($_POST['ID_locale']);
}
$current_date = date('Y-m-d');
$sql = "INSERT INTO recensione (Titolo_R, Voto_R, Commento_R, IDnegozio_R, Email_R, Utente_R, Data_R)
VALUES ('$title_rew', '$voto_rew', '$review_rew','$id_rew','".$_SESSION['emailSessione']."','".$_SESSION['usernameSessione']."','$current_date')";
$result = mysqli_query($conn,$sql) or die(mysqli_error($conn));
//Chiudo la connessione
$conn->close();
header("location:..\locals_page.php");
?>
I've tried with this
$("#submit_review").unbind().click(function() {
var chx = document.getElementsByName("Voto");
for (var i=0; i<chx.length; i++) {
// If you have more than one radio group, also check the name attribute
// for the one you want as in && chx[i].name == 'choose'
// Return true from the function on first match of a checked item
$.post("php/insert_comment.php" );
return true;
}
// End of the loop, return false
alert("Seleziona almeno il voto!");
return false;
});
But there it's not working. It's strange because with a button submit and a
<form role="form" id="review-form" method="post" action="php/insert_comment.php">
It was working. But now I have to do this without a button type "Submit"
Thank you to all in advance
Upvotes: 2
Views: 60
Reputation: 286
Single quotes treat $ as a string
$sql = "INSERT INTO recensione (Titolo_R, Voto_R, Commento_R, IDnegozio_R, Email_R, Utente_R, Data_R)
VALUES ('".$title_rew."', '".$voto_rew."', '".$review_rew."','".$id_rew."','".$_SESSION['emailSessione']."','".$_SESSION['usernameSessione']."','$current_date')";
Upvotes: 0
Reputation: 543
just try this
$.ajax({
url : "php/insert_comment.php",
type : "post",
data : $("#review-form").serialize();
success : function(data){
alert(data); // show when ajax return response from php script
}
})
Upvotes: 2
Reputation: 227
for that registered onclick event on that button and send ajax request using jquery
$.ajax({
type:'POST',
url : "your url",
data : $('form').serializeArray(),
cache: false,
success:function(response){
console.log(response) // your result from php
}
})
Upvotes: 0