I am attempting to write an ajax call that checks against my database and updates a div if it finds changes.
I’m relatively new to JavaScript, and even more new to JSON things.
I think the problem is in how I am returning the JSON array, or how it is used in the if statement, but I’m not sure how to do it properly. I came to that conclusion based on the fact I get a console log return from the success function “Connected and executed PHP page…”, but NOT the “Updated Tickets Page…” log.
Here is my JavaScript:
function checkUpdates()
{
$.ajax({
type: "POST",
url: 'ajax/checkDB.php',
data: {},
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function(data) {
console.log('Connected and executed PHP page...');
if (data.hasChanged == true) {
console.log('Updated Tickets Page...');
$("#contents").load("dynamic/tickets.php");
$("#contents").fadeTo("fast", 1);
}
}
});
}
$(document).ready(function() {
setInterval("checkUpdates()", 3000); // Calls the function every 3 seconds
});
And here is my PHP (checkDB.php, called in the ajax request), if it matters.
<?php
include("../static/config.php");
session_start();
header("Content-Type: text/json");
$result = mysqli_query($con,"SELECT notify FROM tickets WHERE notify='0'");
$row_cnt = mysqli_num_rows($result);
if($row_cnt > 0) {
echo json_encode(array('hasChanged' => 'true'));
mysqli_query($con,"UPDATE tickets SET notify='1' WHERE notify='0'");
} else {
echo json_encode(array('hasChanged' => 'false'));
}
?>
Answer
You are having typo error. You can change either php code without string or change if condition.
//changes made here
if (data.hasChanged == "true") {
PHP
if($row_cnt > 0) {
//here you are having string "true"
echo json_encode(array('hasChanged' => 'true'));
mysqli_query($con,"UPDATE tickets SET notify='1' WHERE notify='0'");
} else {
echo json_encode(array('hasChanged' => 'false'));
}