Pass data from ajax to php [duplicate] - php

Question

This question already has answers here:
How do I return the response from an asynchronous call?
(41 answers)
Closed 7 years ago.
Ajax:
function check_user_country_prod(userId , countryCode , testType )
{ //using alert to check that all data are correct
$.ajax({
type: "POST",
url: "http://localhost/test/testForm.php",
data: { userId: userId ,
countryCode : countryCode ,
productCode: testType
},
success:function(res) {
if(res == "OK")
return true;
else
return false;
}
});
}
PHP:
<?php
require_once("Connections/cid.php");
$userId= $_POST['userId'];
$productCode= $_POST['productCode'];
$countryCode= $_POST['countryCode'];
$sql_check = "SELECT * FROM utc WHERE userId = '$userId' AND productCode = '$productCode' AND countryCode = '$countryCode'";
$list = mysqli_query( $conn, $sql_check);
$num = mysqli_fetch_assoc($list);
if($num >0)
echo "OK";
else
echo "NOK";
?>
I am very sure that the data i had pass in to the php file are correct. However i cant seem to get my data to the php file and it keep return false value back to me. Anything i can do to make it works?
**Note: Somehow even if i change both result to return true in ajax, it will still return false.
and i tried to change the link to another file with only echo "OK"; but it also doesn't work. So i think it is not the file problem. It just never run the ajax no matter what. Do i need to do any link for ajax to run? **

Answer 1

function check_user_country_prod(userId , countryCode , testType )
{ //using alert to check that all data are correct
$.ajax({
url: "test/testForm.php"
type: "POST",
url: "http://localhost/test/testForm.php", // This needs to be "test/testForm.php"
data: { userId: userId ,
countryCode : countryCode ,
productCode: testType
},
success:function(res) {
if(res == "OK")
return true;
else
return false;
}
});
}

Answer 2

Adjust this code according your input type. I hope this will help you:
$(document).ready(function() {
$("#submit_btn").click(function() {
//get input field values
var user_name = $('input[name=name]').val();
var user_email = $('input[name=email]').val();
var user_phone = $('input[name=phone]').val();
var user_message = $('textarea[name=message]').val();
//simple validation at client's end
//we simply change border color to red if empty field using .css()
var proceed = true;
if(user_name==""){
$('input[name=name]').css('border-color','red');
proceed = false;
}
if(user_email==""){
$('input[name=email]').css('border-color','red');
proceed = false;
}
if(user_phone=="") {
$('input[name=phone]').css('border-color','red');
proceed = false;
}
if(user_message=="") {
$('textarea[name=message]').css('border-color','red');
proceed = false;
}
//everything looks good! proceed...
if(proceed)
{
//data to be sent to server
post_data = {'userName':user_name, 'userEmail':user_email, 'userPhone':user_phone, 'userMessage':user_message};
//Ajax post data to server
$.post('contact_me.php', post_data, function(response){
//load json data from server and output message
if(response.type == 'error')
{
output = '<div class="error">'+response.text+'</div>';
}else{
output = '<div class="success">'+response.text+'</div>';
//reset values in all input fields
$('#contact_form input').val('');
$('#contact_form textarea').val('');
}
$("#result").hide().html(output).slideDown();
}, 'json');
}
});
//reset previously set border colors and hide all message on .keyup()
$("#contact_form input, #contact_form textarea").keyup(function() {
$("#contact_form input, #contact_form textarea").css('border-color','');
$("#result").slideUp();
}); });

Answer 3

Try this also both are running on my localhost:
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$(function () {
$('form').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'post',
url: 'post.php',
data: $('form').serialize(),
success: function (d) {
alert(d);
}
});
});
});
</script>
<form method="post" id="myform">
<input type="text" name="time" /><br>
<input type="text" name="date" /><br>
<input name="submit" type="submit" value="Submit">
</form>
make new file of post.php and pest this code. This code will alert your input field value:
<?php print_R($_POST); ?>

Answer 4

"my current folder is localhost/abc/bsd.php while the one that i am going is localhost/test/testForm.php".
From the above comment, you have to change the ajax url to
url: "/test/testForm.php",
Apart from this , your php code shows
$num = mysqli_fetch_assoc($list);
if($num >0)
echo "OK";
This is incorrect as mysqli_fetch_assoc returns an associative array of strings. So, the comparison $num >0 is illogical.