PHP Call with Ajax

Question

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

first of all your code is working on php or ajax? i think $.post("php/insert_comment.php" ); return true; is not working because on submit of your form action attribute redirect your data to that url then I think your click event not getting worked — KARAN LAGALWAR
– KARAN LAGALWAR, Commented Aug 4, 2017 at 10:37

Mahipal Patel · Accepted Answer · 2017-08-04 10:33:14Z

2

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
    }
})

answered Aug 4, 2017 at 10:33

Mahipal Patel

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KARAN LAGALWAR · Accepted Answer · 2017-08-04 10:32:42Z

0

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
}
})

answered Aug 4, 2017 at 10:32

KARAN LAGALWAR

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Comments

Jasbeer Rawal · Accepted Answer · 2017-08-04 10:46:43Z

0

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')";

answered Aug 4, 2017 at 10:46

Jasbeer Rawal

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