Ajax code for the counterpart of success - php

Question

I have this code to do autocomplete function from database. what if the input is not in database, how can i show the error message?
this is my js code:
$('#txt_fname').autocomplete({
source: function( request, response ) {
$.ajax({
url : 'under the hood/rehabCreateAjax.php',
dataType: "json",
method: 'post',
data: {
name_startsWith: request.term,
type: 'country_table',
row_num : 1
},
success: function( data ) {
response( $.map( data, function( item ) {
var code = item.split("|");
return {
label: code[0],
value: code[0],
data : item
}
}));
}
});
},
autoFocus: true,
minLength: 0,
select: function( event, ui ) {
var names = ui.item.data.split("|");
$('#txt_mname').val(names[1]);
$('#txt_sname').val(names[2]);
$('#txt_nickname').val(names[3]);
$('#txt_stuNum').val(names[4]);
}
});
this is my php code:
require_once '../connection/dbConn.php';
if($_POST['type'] == 'country_table'){
$row_num = $_POST['row_num'];
$name = $_POST['name_startsWith'];
$query = "SELECT stu_fname, stu_mname, stu_sname, stu_nickname, student_id FROM student
where (UPPER(stu_fname) LIKE '".strtoupper($name)."%') or (UPPER(stu_mname) LIKE '".strtoupper($name)."%') or
(UPPER(stu_sname) LIKE '".strtoupper($name)."%') or (UPPER(stu_nickname) LIKE '".strtoupper($name)."%')";
$result = mysqli_query($conn, $query);
$data = array();
while ($row = mysqli_fetch_assoc($result)) {
$name = $row['stu_fname'].'|'.$row['stu_mname'].'|'.$row['stu_sname'].'|'.$row['stu_nick name'].'|'.$row['student_id'].'|'.$row_num;
array_push($data, $name);
}
echo json_encode($data);
}
by the way..i have tried to put else in my php code.. not working.

Answer 1

If I understood correctly you want to respond to an empty results set, and not to some sort of server error. If this is the case you should include it in your success callback. simply check in your client code if the data array has items and respond accordingly:
success: function( data ) {
if (data.length) {
response( $.map( data, function( item ) {
var code = item.split("|");
return {
label: code[0],
value: code[0],
data : item
}
}));
} else {
console.log("no results");
}
}
The error callback suggested by #Koogle is not meant for this purpose, and you definitely should not return a 404 from the server, which is only used when the url itself is not found (or for that matter a 500, which implies an internal server error).
The error callback represents a low-level failure to get a response from the server, and is automatically ran by JQuery if the request failed. It should not be forced to run by returning an error code for a successful request (200). The error you are referring to is high-level, because it stems from the inherent logic of your code, and not from an objective failure to execute the code itself for whatever reason.
Thus, the label error in your case is dictated by the way you perceive the logic of your code. e.g if you are filtering results in a table and the filter doesn't return any results, this can hardly qualify as an error. However, if you expect that the request would always return at least one result, then error may be justified, although I believe that exception is the more accurate term.

Answer 2

You could respond in your php script with an HTTP error message and handle the error in JS as follows:
$.ajax({
url : 'under the hood/rehabCreateAjax.php',
dataType: "json",
method: 'post',
data: {
name_startsWith: request.term,
type: 'country_table',
row_num : 1
},
success: function( data ) {
response( $.map( data, function( item ) {
var code = item.split("|");
return {
label: code[0],
value: code[0],
data : item
}
}));
error: function (response) {
// Handle error here
console.log(response);
}
}
The PHP should be something like:
$query = "SELECT stu_fname, stu_mname, stu_sname, stu_nickname, student_id FROM student
where (UPPER(stu_fname) LIKE '".strtoupper($name)."%') or (UPPER(stu_mname) LIKE '".strtoupper($name)."%') or
(UPPER(stu_sname) LIKE '".strtoupper($name)."%') or (UPPER(stu_nickname) LIKE '".strtoupper($name)."%')";
$result = mysqli_query($conn, $query);
$data = array();
while ($row = mysqli_fetch_assoc($result)) {
$name = $row['stu_fname'].'|'.$row['stu_mname'].'|'.$row['stu_sname'].'|'.$row['stu_nick name'].'|'.$row['student_id'].'|'.$row_num;
array_push($data, $name);
}
// Respond with error
if($data.length() == 0) {
http_response_code(500);
echo "No such entry";
return;
}
echo json_encode($data);