PHP/jQuery Instant Search Not Working

Question

I'm trying out a tutorial to learn how to do instant search with PHP/jQuery. I can't seem to find why this code won't work. This search was working before when I had the PHP in the same file as the index, but when I moved it to another file, it stopped working. I keep getting this console error message with each keystroke: ReferenceError: Can't find variable: $_POST. Any help would be deeply appreciated.

index.php file

<!DOCTYPE html>
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset-utf-8">
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
    <script type="text/javascript">
        function searchkey() {
            var searchTxt = $("input[name='search']").val();
            $_POST("search.php", {searchVal: searchTxt}, function(output) {
                $("#output").html(output);
            });
        }
    </script>
    <title>Search</title>
</head>
<body>
    <form action="index.php" method="post">
        <input type="text" name="search" placeholder="Search for members..." onkeyup="searchkey();">
        <input type="submit" value="Search">
    </form>
    <div id="output"></div>
</body>
</html>

search.php file (same location as index.php)

<?php
$connection = mysqli_connect('localhost','root','root','LBD');
$output='';

if(isset($_POST['searchVal'])){
    $searchkey= $_POST['searchVal'];
    $searchkey=preg_replace("#[^0-9a-z]#i", "", $searchkey);

    $query = mysqli_query($connection,"SELECT * FROM members WHERE ownerName LIKE '%$searchkey%' OR companyName LIKE '%$searchkey%'") or die("Could not search!");
    $count = mysqli_num_rows($query);

    if($count == 0){
        $output="There was no search result!";
    }
    else{
        while($row=mysqli_fetch_array($query)){
            $oName=$row['ownerName'];
            $cName=$row['companyName'];

            $output .='<div>'.$oName.'<br/>'.$cName.'</div>';
        }
    }
}
echo ($output);
?>

Did you mean to use jQuery's shorthand post method: $.post() which is different from $_POST which is the super global variable in PHP — Patrick Evans
– Patrick Evans, Commented Nov 9, 2017 at 3:27
Thank you @PatrickEvans that did the trick! I really appreciate the help :) — Sam Sverko
– Sam Sverko, Commented Nov 9, 2017 at 3:50

giolliano sulit · Accepted Answer · 2017-11-09 03:28:21Z

1

Looks like you've used the PHP $_POST in your script..

Try to use:

$.POST

answered Nov 9, 2017 at 3:28

giolliano sulit

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1 Comment

Sam Sverko Over a year ago

That worked! Thank you so much. This was driving me crazy. I really appreciate the help :)

guradio · Accepted Answer · 2017-11-09 03:38:07Z

0

Try this

$.ajax({
    url: "search.php",
    type: 'POST',
    data: {
      searchVal: searchTxt
    },
  })
  .done(function(output) {
    $("#output").html(output);
  });

edited Nov 9, 2017 at 3:38

guradio

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answered Nov 9, 2017 at 3:33

shohidul

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