Retrieving multiple values via PHP/jQuery/Ajax - php

Question

I'm using the following jQuery to retrieve values for a 'Live Search' field:
$(document).ready(function(){
/* LIVE SEARCH CODE - START HERE*/
var UserID = ('<?php echo $_SESSION['UserID'] ?>');
$(document).ready(function(){
$('.clLiveSearchAccount').on("keyup" , "[id*=txtAccountID]", "input", function(){
/* Get input value on change */
var inputVal = $(this).val();
var ParentTransID = $(this).prev().val();
alert(UserID);
var resultDropdown = $(this).siblings(".result");
if(inputVal.length){
$.get("Apps/Finance/incGetAccounts.php", {term: inputVal, usr: UserID }).done(function(data){
// Display the returned data in browser
resultDropdown.html(data);
});
} else{
resultDropdown.empty();
}
});
// Set search input value on click of result item
$(document).on("click", ".result p", function(){
$(this).parents(".clLiveSearchAccount").find('#txtAccountID').val($(this).text());
$(this).parent(".result").empty();
});
});
I'm using this PHP ajax handler:
<?php
/* ------------------------------------------------ */
$link = mysqli_connect("xxx", "xxx", "xxx", "xxx");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security
$term = mysqli_real_escape_string($link, $_REQUEST['term']);
$user = mysqli_real_escape_string($link, $_REQUEST['usr']);
if(isset($term)){
// Attempt select query execution
$sql = "SELECT * FROM tblAccounts WHERE Name LIKE '%" . $term . "%' AND UserID=" . $user;
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
echo "<p>" . $row['Name'] . "</p>";
}
// Close result set
mysqli_free_result($result);
} else{
echo "<p>No matches found</p>";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
}
// close connection
mysqli_close($link);
?>
But how do I send back (and accept) an additional value, so the String Name and Integer Key?

Answer 1

It seems you are looking to send JSON data. Put the HTML you want to echo in a variable.
$html ="<h1>PHP is Awesome</h1>";
$myVariable = 3;
echo json_encode(array( 'variable1' => $myVariable, 'html' => $html ));
and you'll need a success callback in your javascript
success: function($data) {
var html = $data.html;
var myVar = $data.variable1;
// other stuff
}
look up a tutorial on PHP JSON
W3schools always a good place to start
https://www.w3schools.com/js/js_json_php.asp

Answer 2

I always do this return format in ajax.
Success Response
// PHP
$result = [
'success' => true,
'variable' => $myVariable,
'html' => $html,
];
Fail Response
// PHP
$result = [
'success' => false,
'err_message' => 'Error message here!',
],
use json encode when returning the data to the ajax example json_encode($result) and also dont forget to add dataType setting in your ajax so that it will expect json format response.
Ajax fn
$.ajax({
url: 'path to php file',
type: '' // specify the method request here i.e POST/GET
data: {} // data to be send
dataType: 'JSON',
success: function(res) {
if (res.success) {
...
} else {
// you can put the error message in html by accessing res.err_message
}
}
});