Compare user value to database and show result through ajax jquery - php

Question

Guys m working on my first live project and i am stuck at a point, where i need help with ajax jquery. i can do this with PHP but i wanna do this with ajax.
Here if user enter a product code ,so i want to compare this product code value into my database and show product name in my other form ,which will open after user input value:
Here in first field i want product name:
Here in my table you can see product code and product name:
ok so here is my html code in last option when user enter product code
Here is jquery i am sending user data to 8transectiondata.php to compare
And this is php file and i want $data['product_name']; to show

Answer 1

Here's a generic answer.
JS FILE:
$(document).ready(function () {
$('#myButtonId').on('click', function () {
var code = $('#myCodeInputId').val();
if (code !== '') { // checking if input is not empty
$.ajax({
url: './my/php/file.php', // php file that communicate with your DB
method: 'GET', // it could be 'POST' too
data: {code: code},
// code that will be used to find your product name
// you can call it in your php file by "$_GET['code']" if you specified GET method
dataType: 'json' // it could be 'text' too in this case
})
.done(function (response) { // on success
$('#myProductNameInput').val(response.product_name);
})
.fail(function (response) { // on error
// Handle error
});
}
});
});
PHP FILE:
// I assumed you use pdo method to communicate with your DB
try {
$dbh = new PDO('mysql:dbname=myDbName;host=myHost;charset=utf8', 'myLogin', 'myPassword');
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $e) {
exit('ERROR: ' . $e->getMessage());
}
$sql = "SELECT `product_name` FROM `products` WHERE `product_code` = :code";
$result = $dbh->prepare($sql);
$result->bindValue('code', $_GET['code'], PDO::PARAM_INT);
$result->execute();
if($result->rowCount()) { // if you got a row from your DB
$row = $result->fetchObject();
echo json_encode($row, JSON_UNESCAPED_UNICODE); // as we use json method in ajax you've got to output your data this way
// if we use text method in ajax, we simply echo $row
}
else {
// handle no result case
}

Answer 2

I know what you want to do, but without specific code the best I can do is give you a generalized answer.
When a user fills out a field, you want to post that field to the server, look up a product and return some stuff.
The basics are going to look like this.
$(document).ready( function(){
//rolling timeout
var timeout;
$('#field').on('keyup', function(e){
if(timeout) clearTimeout(timeout);
timeout = setTimeout( function(){
var data = {
"field" : $('#field').val()
};
$.post( '{url}', data, function(response){
if(response.debug) console.log(response.debug);
if(response.success){
//open other form
$('{otherFormProductField}').val(response.product);
}
}); //end post
},450); //end timeout
});//end onKeyup
}); //end onReady
Then in PHP, you have to process the request. Pull the field from the $_POST array, look it up in the Database. Then build a response array and send it back to the client as JSON. I like to build responses in a structure something like this.
{
success : "message", //or error : "message"
debug : "",
item : ""
}
Then in PHP I will do this.
ob_start();
..code..
$response['debug'] = ob_get_clean();
header("Content-type:application/json");
echo json_encode($response);
This way, you can still print out debug info (in side the output buffer calls ) when developing it and don't have to worry about it messing up the Json or the header call.
-note- Use a timeout, that you reset on each key press (a rolling timeout). What it does is reset the previous timeout each time the key is released. That way it only sends the request once the user quits typing (instead of sending request on every keypress). I have found 450 milliseconds to be about the perfect value for this. Not too long not too short. Basically once they stop typing for 450ms it will trigger the $.post