marky marky - 4 years ago 60
jQuery Question

Can't figure out why PHP not receiving POST data from $.ajax call

It's not like I haven't done this same process before, but I can't figure out why my PHP script's POST data is empty. Here's what I've done/found:

  1. I've verified that the $.ajax call's "data" parameter has a value (alerts in the submitSearch function and in the success parameter show the correct value of the search variable).

  2. I know that the script is being "found" - no file not found messages in the js console in the browser.

  3. I'm not getting any db connection errors

  4. I also know that the PHP script is running because the alert in the $.ajax call's success parameter is displaying the $message value in the PHP script's else clause.

  5. The logging I have set up in the PHP script is displaying nothing for the POST data

I'd greatly appreciate it if someone can take a look at this and hopefully point out what I'm missing.

Here is all of the relevant code:

Javascript (in a script tag within the HTML file)

$('input#btnSubmitSearch').click(function() {
// Clear the text box in case an error was indicated previously
$('input#txtSearch').css({'background-color' : ''});

var search = $('input#txtSearch').val();

if (search == '' || search.length != 5) {
alert ('Not a valid entry!');
$('input#txtSearch').css({'background-color' : '#FDC3C3'});
return false;
else {
return false;

function submitSearch(search) {
alert ('Sending ' + search + ' to the database.');

type: 'POST',
url: 'scripts/search.php',
data: search,
cache: false,
success: function(response) {
alert ('search is: ' + search + ', Response from PHP script: ' + response);
error: function(xhr) {
var response = xhr.responseText;
var statusMessage = xhr.status + ' ' + xhr.statusText;
var message = 'Query failed, php script returned this status: ';
var message = message + statusMessage + ' response: ' + response;

PHP script (scripts/search.php)

require_once ('logging.php');
$log = new Logging();

$dbc = @mysqli_connect([connection stuff])
OR die ('Could not connect to MySQL server: ' . mysqli_connect_error() );

$log->lwrite('$_POST[\'search\']: ' . $_POST['search']);

if (isset($_POST['search'])) {

$search = $_POST['search'];

$log->lwrite('$search: ' . $search);

$querySearch= "SELECT id, value
FROM table
WHERE value LIKE '%" . $search . "%'";

$log->lwrite('$querySearch: ' . $querySearch);

$resultSearch = @mysqli_query($dbc, $querySearch);

$numRowsSearch = mysqli_num_rows($resultSearch);

$log->lwrite('rows returned: ' . $numRowsSearch);

if ($numRowsSearch > 0) {
while ($rowSearch = mysqli_fetch_array($resultSearch, MYSQLI_ASSOC)) {
echo 'Value found: ' . $rowSearch['value'];
else {
$errorMessage ='$_POST[\'search\'] doesn\'t have a value!';
echo ($errorMessage);

Answer Source
    type: 'POST',
    url: 'scripts/search.php',
    data: {'search': search}, 

I believe you need to set $_POST['search'] as above.

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