如何将PHP变量与Typeahead变量一起传递

Question

I have cut this down to be a simple as possible. I create a typeahead variable that works perfectly. but I need to pass two other variables $php_var1 and $php_var2 that are unrelated to the typeahead. The PHP variables are defined in start.php. The typeahead script calls search_script.php then calls cart.php. cart.php is were I will need the two PHP variables to be passed to. Thanks in advance for any help

start.php

<?php
  $php_var1 = "my php variable 1";
  $php_var2 = "my php variable 2";
?>


<script>

$(document).ready(function() {

    var php_var1 = <?php echo $php_var1; ?>;
    var php_var2 = <?php echo $php_var2; ?>;


    $('#my_input').typeahead({
        source: function(query, result) {
            $.ajax({
                url: "search_script.php",
                method: "POST",
                data: {
                    query: query
                },
                dataType: "json",
                success: function(data) {
                    result($.map(data, function(item) {
                        return item;

                    }));
                }

            })
        },
        updater: function(item) {
            location.href = 'cart.php?shop_name=' + item
            return item
        }
    });

    });

</script>

<form action="cart.php" action="post">

<input type="text"  id="my_input" placeholder="Typeahead Search" />

</form>

search_script.php

<?php

$php_var1  = isset($_REQUEST['php_var1']) ? $_REQUEST['php_var1'] : "empty";
$php_var2  = isset($_REQUEST['php_var2']) ? $_REQUEST['php_var2'] : "empty";

$connect = mysqli_connect($servername, $username, $password, $dbname);
$request = mysqli_real_escape_string($connect, $_POST["query"]);
$query = " SELECT * FROM all_shops WHERE p_shop_name LIKE '%".$request."%'";

$result = mysqli_query($connect, $query);

$data = array();

if(mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_assoc($result))
{
$data[] = $row["p_shop_name"];
}
echo json_encode($data);

}
?>

cart.php

$php_var1  = isset($_REQUEST['php_var1']) ? $_REQUEST['php_var1'] : "empty";
$php_var2  = isset($_REQUEST['php_var2']) ? $_REQUEST['php_var2'] : "empty";

echo $php_var1;
echo $php_var2;

?>

Answer 1

You need quotes around the php output in order to generate javascript strings

var php_var1 = "<?php echo $php_var1; ?>";
var php_var2 = "<?php echo $php_var2; ?>";

Answer 2

Stackoverflow is an excellent resource, but sometimes you don't get the answer, so you need to persevere and keep trying. I worked on this all day yesterday and just couldn't figure it out. Woke up this AM and it came to me. The answer is as follows. In the typeahead script change the following line

location.href = 'cart.php?shop_name=' + item      

to

location.href = 'cart.php?shop_name=' + item + '&php_var1=<?php echo $php_var1 ?>' + '&php_var2=<?php echo $php_var2 ?>'