ajax is not working with jquery - php

Question

Here is my ajax code.
I cannot see any error in my code, but AJAX doesn't work.
It doesn't return anything from that page...
function addCash(){
var cash =$('#cash_amount').val();
var date =$('#cash_date').val();
var debiter =$('#debiter').val();
if(cash == '' || date =='' ){
alert("Please Fill All Fields");
}
else{
$.ajax({
type: 'POST',
dataType:'JSON',
url: 'getCustomers.php',
data: 'type=cash_received&cash='+cash+'&date='+date+'& debiter='+debiter,
success:function(data){
console.log(data);
alert("Cash Added Successfully");
}
});
}
}
PHP Code "getCustomers.php"...inside a function using ajax is an issue?
$cash= $_REQUEST['cash'];
$date= $_REQUEST['date'];
$debiter= $_REQUEST['debiter'];
$query="INSERT INTO `received_payment`(`debiter`, `amount`, `date`) VALUES ('".$debiter."', '".$cash."', '".$date."')";
$result = $mysqli->query($query) or ($error=$mysqli->error.LINE);
$arr = array();
if(!$result){
$arr['result'] = isset($error) ? $error : 'error';
}
else{
$arr['result'] ="ok";
}
$json_response = json_encode($arr);
ob_clean();
echo $json_response;`

Answer 1

Because, you are using die anf if your query fails then your script will die and hence no response will be made. So change the below line,
$result = $mysqli->query($query) or die($mysqli->error.__LINE__);
to
$result = $mysqli->query($query) or ($error=$mysqli->error.__LINE__);
and you can return this error in response like,
if(!$result){
$arr['result'] = isset($error) ? $error : 'error';
}
Also in your insert query, fields and their values are not matching, you should use it like,
$query="INSERT INTO `received_payment` (`debiter`, `amount`, `date`)
VALUES ('".$debiter."', ".$cash."', '".$date."')";
And try to pass data from AJAX (you have space before debiter key in your data string) like,
data: {type:'cash_received',cash:cash,date:date,debiter:debiter},

Answer 2

An extra comma in INSERT query.
$query="INSERT INTO `received_payment`(`debiter`, `amount`, `date`,) VALUES ('".$cash."', '".$date."', '".$debiter."')";
Change to:
$query="INSERT INTO `received_payment`(`debiter`, `amount`, `date`) VALUES ('".$cash."', '".$date."', '".$debiter."')";
It is giving an error actually, but since you gave die() the code is getting exited. So you are not getting anything from Server.

Answer 3

in AJAX request you have mentioned accept only JSON data content. So in some case it may happens server returns response with Warning and Error if PHP error is on.
So in server side PHP script before echo response in json format you can use ob_clean() for flushing all unexpected output which is sent by error or warning.
$json_response = json_encode($arr);
ob_clean();
echo $json_response;

Answer 4

Please remove quotes at the end of the line
echo $json_response;`